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Applied optimization problems.
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Example.
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The rectangle is inscribed in the region in the first quadrant bound by the coordinate axes and the parabola y equals 1 minus x squared.
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Find the dimensions of the rectangle that maximize its area.
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Let's begin with a sketch of the part of the parabola that we're interested in. We can assume in order for a rectangle inscribed in this region to have a maximum area,
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two of its sides will be on the coordinate axes and its upper right corner will touch the parabola.
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Here's one such rectangle whose area is probably less than the maximum, as is this one.
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This rectangle and this one have larger areas and the rectangle with maximum area probably looks something like this one.
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Now let's label the corner that touches the parabola with coordinates x comma 1 minus x squared.
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Notice that the width of the rectangle is the x coordinate at that point and its height is the y coordinate 1 minus x squared.
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So the area of the rectangle width times height is simply x times y, where y is equal to 1 minus x squared.
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By substituting y equals 1 minus x squared in the area formula, we obtain the area as a function of x alone, a of x equals x times 1 minus x squared, or x minus x cubed.
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The domain of interest for this function is the closed interval 0, 1.
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So we now have posed a minimization problem for a continuous function on a closed bounded interval.
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To find critical numbers, we begin by taking the derivative, a prime of x is 1 minus 3 x squared, which is 0 at x equal to the cube root of 1 third.
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So we have a single critical number in the interval 0 to 1.
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Since the domain is a closed bounded interval, we'll compute the value of a at the critical number, x times 1 minus x squared, when x equals the square root of 1 third, turns out to be 2 over 3 root 3, which is approximately 0.385.
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Then we compute the endpoint values, a of 0 equals 0, and a of 1 equals 0 also. Therefore, the maximum value of a is a of the square root of 1 third equals 2 over 3 root 3.
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Here's the graph of x minus x cubed. Here, we're only concerned with the part corresponding to the interval 0, 1, where the endpoint values are 0, and the maximum on the interval occurs at the single critical number inside that interval.
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Finally, since the problem asks for the dimensions of the largest rectangle, we state the final answer by reporting the optimal dimensions.
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The width x is square root of 1 third, and the height y is 1 minus x squared, or 2 thirds.
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Now let's look back and highlight the essential steps in the solution. We began with a figure labeled with relevant variables.
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Next, we wrote down a formula, here, a equals x times y, for the quantity whose maximum we hope to find. This is our objective function.
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Since the objective function involved two variables, we had an additional equation called a constraint, which came from the fact that one corner of the rectangle is constrained to the parabola.
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The constraint provided a relationship between the two variables that allowed us to then express the objective function in terms of just one variable.
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We then computed the derivative of that function and solved for its critical numbers.
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Next, we did an analysis of the critical number to determine whether our function contained a maximum there.
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Since the domain was a closed bounded interval, we simply compared the value of the function at the critical number and the endpoint values.
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Finally, we stated an answer to the problem as posed. We will see that these are common elements in the solution of many optimization problems.
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Example. Find the point on the graph of y equals 1 over x squared, x greater than 0, that is closest to the origin.
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Here is the graph of y equals 1 over x squared. Notice that there is obviously a point on this curve that is closest to the origin.
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Let's label a generic point x comma 1 over x squared.
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The first key observation here is a simple one. The point on the curve that is closest to the origin is the one that minimizes the distance to the origin.
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Using the distance formula, we state our objective function. It's the square root of x squared plus y squared.
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The point xy is constrained to the curve y equals 1 over x squared, so that equation is the constraint that relates x and y.
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Substituting y equals 1 over x squared into the objective function gives us the square root of x squared plus 1 over x to the 4th.
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Now we make another observation that will simplify our job. We may as well work with the square of the distance, since it will be a minimum at the same number that the distance itself is a minimum.
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So let's define f of x equals x squared plus 1 over x to the 4th, which we can also write as x to the 6th plus 1 over x to the 4th.
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Our domain for this function is all x greater than 0.
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So we have a minimization problem posed on an open, unbounded interval.
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Now let's compute the derivative. f prime of x equals 2x minus 4 times x to the minus 5 power, which we can also write as 2x to the 6 minus 4 over x to the 5th.
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So f prime of x is equal to 0 at x equal to the 6th root of 2, which is approximately 1.12.
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So we have a single critical number in the domain of the function. To analyze this critical number, let's look at the second derivative, which here is 2 plus 20 times x to the minus 6 power, which is clearly positive for all x greater than 0.
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Therefore, the function attains its minimum value at the critical number, the 6th root of 2.
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And the value there is 3 times the cube root of 2 over 2.
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Remember, this is the square of the minimum distance. So the minimum distance is its square root, which is approximately 1.37.
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Finally, since the problem asks for the point on the curve, we provide that as our final answer. The x coordinate is the 6th root of 2, and the y coordinate is 1 over that squared, which is 1 over the cube root of 2.
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Example, an aquarium is to be built to hold 20 cubic feet of water. If 2 ends of the aquarium are square, and there's no top, find the dimensions that minimize the surface area, and thus the amount of glass used in its construction.
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Let's start with a simple figure. We have a box with no top, the front and back ends here are square.
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Let's let x be the width of the base, as well as the height, since the ends are square, and let's let l be the length of the aquarium.
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Since we want to minimize the surface area, that would be our objective function.
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The area of the 2 square ends combined is 2 times x squared.
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The front, back, and bottom of the aquarium, each have area x times l, and so we have 3 times x times l, accounting for the area of those 3 sides.
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Our objective function is a equals 2x squared plus 3xl. The constraint that links the 2 variables x and l, is simply the volume requirement. The volume of the aquarium is x squared times l, and so we have x squared times l equals 20, which we can easily solve for l to obtain l equals 20 over x squared.
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So now we can write down the area as a function of x alone by substituting l equals 20 over x squared into the formula.
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So we have 2x squared plus 60 over x. The domain of this function is all x greater than 0.
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So let's calculate the derivative, a prime of x, is equal to 4x minus 60 times x to the minus 2 power, or 4x minus 60 over x squared, which we can write over a common denominator as 4 times x cubed minus 15 over x squared.
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This form makes it clear that a prime of x is equal to 0, and x equal to the cube root of 15, which is approximately 2.47. So we have a single critical number in the domain of the function.
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Since we're working on an open unbounded interval, let's calculate the second derivative, which will be 4 plus 120 times x to the minus 3 power, which is clearly positive for all x in the domain of our function.
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Therefore, we conclude that the graph is always concave up. In fact, it looks like this graph here. And so, a, it contains a minimum value at the critical number, cube root of 15, and that value is 2 times the cube root of 15 squared plus 60 over the cube root of 15.
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Noting that the second term here is the same as 4 times the cube root of 15 squared, we can simplify this to obtain 6 times the cube root of 15 squared, which is approximately 36.49 square feet.
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Now, since the problem asks for the dimensions that minimize the surface area, we report the optimal dimensions here, x equals the cube root of 15, which is approximately 2.47 feet.
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And the length using the constraint equation is 20 over x squared, so we have 20 over 15 to the 2,3 power, or 4,3 times the cube root of 15, which is simply 4,3 times x. This is approximately 3.23 feet.
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Here's another example that's actually quite similar to the previous one. The juice can, in the shape of a right circular cylinder, is to have a volume of 1 liter or 1000 cubic centimeters.
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Find the height and radius that minimizes the surface area of the can, and thus the amount of material used in its construction.
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Here's a picture of a typical right circular cylinder. We've labeled the radius r and the height h. We want to minimize the surface area, so our objective function is the surface area.
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The top and the bottom sides of the can, each have area pi r squared, so together they give us 2 pi r squared. The vertical surface of the can has area equal to the circumference of the rim 2 pi r times the height h, so our area is 2 pi r squared plus 2 pi r h.
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Our constraint is the fact that the volume is to be 1 liter. The volume of the can is the area of the base pi r squared times the height h, so we have pi r squared h equals 1000 cubic centimeters.
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So let's solve for h and write h equal to 1000 over pi r squared. Now, we'll substitute our constraint into the area equation and obtain a function of 1 variable r, that's 2 times the quantity pi r squared plus 1000 r, with the domain of all r greater than 0.
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Now, let's take the derivative, a prime of r is 2 times 2 pi r minus 1000 times r to the minus 2 power. Now, let's factor out a 2 and write the expression over a common denominator of r squared. The numerator then is pi r cubed minus 500, which makes it clear that a prime of r is equal to 0 at r equal to the cube root of 500 over pi.
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Since 500 is 125 times 4 or 5 cubed times 4, we can write this as 5 times the cube root of 4 over pi.
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So we see again that we've got a single critical number in the domain of the function.
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Since that domain is an unbounded open interval, let's look at the second derivative, a double prime of r, that will be 2 times 2 pi plus 2000 times r to the minus 3 power, which is clearly positive for all positive r.
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So we conclude that the graph of a is always concave up, and so a must attain its minimum value at the critical number. The minimum area is the value of a at 5 times the cube root of 4 over pi, which turns out to be 300 times the cube root of 2 pi, which is approximately equal to 554 square centimeters.
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The problem asks for the height and the radius that minimizes the surface area of the can, so we'll give the optimal dimensions.
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The radius should be 5 times the cube root of 4 over pi, or approximately 5.42 centimeters.
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In the corresponding height, we calculate from the constraint equation, h equals 1000 over pi r squared, which simplifies to 10 times the cube root of 4 over pi, unless notice that that is exactly 2 times the optimal radius.
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This tells us that the surface area of the can is minimized when the height is equal to the can's diameter, so the can with minimal surface area will actually have a square vertical cross section.
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One last example.
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Find the dimensions of the cone with minimum volume that can contain a sphere with radius r. Let's look at a few pictures to get a feel for what we're after here.
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Here's a picture of the sphere inside a cone. Notice that for its particular shape, the cone's volume is minimal since it fits tightly around the sphere.
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With a large radius and a height close to the diameter of the sphere, the volume is large. With a large height and a radius close to the radius of the sphere, its volume is also large.
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So we expect that somewhere between the extremes, the cone will have a minimum volume.
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Let's let h be the height of the cone and little r its radius. Remember that capital r is the radius of the sphere. Since we want to minimize the volume of the cone, our objective function will be the cone's volume, 1-3 pi r squared h.
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Since that involves two unknowns, r and h, we need a constraint to relate to.
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That constraint must come from the requirement that the cone fits snugly around the sphere. To find it, let's look at the central vertical cross-section.
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The line through the middle divides the figure into a pair of identical right triangles. Each of those has a base of length little r and a hypodinous of length square root of h squared plus r squared.
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The vertical leg consists of a radius of the circle with length capital r and a segment with length h minus capital r.
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Now the key observation. The radius of the circle that extends to the point where the circle touches the hypodinous of either right triangle is perpendicular to that side and therefore shares the same angles as the larger right triangle in which it's drawn.
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The larger triangle is shaded here in red and the new similar triangle is shaded here in yellow.
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So we can use the principle of similar triangles to write the appropriate constraint.
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We begin with a smaller triangle. The ratio of the type hypodinous to its shortest leg is h minus capital r over capital r.
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And the larger triangle, the ratio of hypodinous to shortest leg, is the square root of h squared plus a little r squared over a little r.
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These ratios are equal because the triangles are similar.
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Our goal is to reduce the volume formula to one involving only r or only h. To do this, we could solve the constraint equation for r or we could solve it for h.
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Or we could solve it for r squared since r appears in the volume formula only in the form of r squared.
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The algebra involved will be simplest if we just solve for r squared.
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Let's first cross multiply and then square both sides.
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Next, we multiply out the right side and subtract the term capital r squared, little r squared from both sides, and factor little r squared from the left side.
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The other factor on the left simplifies to h squared minus 2rh.
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Now let's simplify by dividing both sides by h.
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And finally, dividing both sides by h minus 2r gives us r squared equal to capital r squared times h divided by h minus 2r.
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So now we substitute that expression for r squared into the volume formula to obtain the volume as a function of h alone.
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Its domain is all h greater than 2r, that is, all h greater than the diameter of the sphere.
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Next, we compute the derivative used in the quotient rule.
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V prime of h is equal to the constant pi r squared over 3 times the quotient whose denominator is the square of h minus 2r, and whose numerator is the derivative of h squared to h times h minus 2r.
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Minus h squared times the derivative of h minus 2r, which is just 1.
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Now we simplify the numerator.
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It becomes h squared minus 4 capital r h.
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We can factor that to get h times h minus 4r in the numerator.
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This makes it easy to see that the prime of h is equal to 0 when h equals 4r. That's the only critical number in the domain of function.
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Now let's justify the claim that v attains a minimum value with this critical number as follows.
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First, let's note that v of h is a continuous function on the interval from 2r to infinity, because it's a rational function of h whose denominator is never 0 on that interval.
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Also, the limit as h approaches 2r from the right of v of h is equal to plus infinity, because the numerator and denominator are both positive and the denominator approaches 0.
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Also, the limit as h approaches infinity of v of h is equal to plus infinity, because the numerator of the rational function has a greater degree than the denominator.
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Therefore, v of h must attain a minimum value somewhere, and since there's only one critical number, the minimum must be attained there.
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So our minimum volume is v of 4r, which turns out to be 8 pi r cubed, which is exactly 2 times the volume of the sphere.
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And since we're asked to find the dimensions of the cone with minimum volume, we'll report the optimal dimensions, 8 equals 4r, and substituting that into the formula for little r squared above, we find that little r squared is equal to 2 times big r squared, in other words, the radius little r is equal to square root of 2 times the radius of the sphere.
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Then one of the clearly-st predictive dimensions is wiggle slope.
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In particular, this is the exact size-wise bigger class-wise giving therefore triple kilowatt-hour evaluation for the Newton interval, oh laughs
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You