Math 181 Miniproject 8: Applied Optimization.md
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tags: MATH 181
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Math 181 Miniproject 8: Applied Optimization
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**Overview:** This miniproject focuses on a central application of calculus, namely *applied optimization*. These problems augment and extend the kinds of problems you have worked with in WeBWorK and class discussions.

**Prerequisites:** Section 3.4 of *Active Calculus.*

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For this miniproject, select EXACTLY TWO of the following and give complete and correct solutions that abide by the specifications for student work. Include a labeled picture with each solution. Full calculus justification of your conclusions is required.

**Problem 1.** Two vertical poles of heights 60 ft and 80 ft stand on level ground, with their bases 100 ft apart. A cable that is stretched from the top of one pole to some point on the ground between the poles, and then to the top of the other pole. What is the minimum possible length of cable required? Justify your answer completely using calculus.
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The following figure shows an estimation of what the problem looks like.
We know that the total length of cable $L$ consists of $L_1+L_2$. We want to find the $x$ value that will minimize this function $L$. Using the Pythagorean Theorem we find that $L_1=\sqrt{3600+x^{2}}$ and $L_2=\sqrt{x^{2}-200x+10000}$
$L=\sqrt{3600+x^{2}}+\sqrt{x^{2}-200x+10000}$
$L'=\frac{1}{2}(\frac{2x}{\sqrt{3600+x^{2}}})+\frac{1}{2}(\frac{2x-200}{\sqrt{x^{2}-200x+10000}})$
$L'$ is defined at ever point. Now we set $L'$ equal to zero.
$0=\frac{1}{2}(\frac{2x}{\sqrt{3600+x^{2}}})+\frac{1}{2}(\frac{2x-200}{\sqrt{x^{2}-200x+10000}})$
$0=(\frac{x}{\sqrt{3600+x^{2}}})+(\frac{x-100}{\sqrt{x^{2}-200x+10000}})$
$0=(\frac{x}{\sqrt{3600+x^{2}}})+(\frac{x-100}{x-100})$
$-1=(\frac{x}{\sqrt{3600+x^{2}}})$
$-\sqrt{3600+x^{2}}=x$
$-(3600+x^2)=x^2$
$-3600=2x^2$
$\sqrt{1800}=x$
The other possible critical values of x are 0 and 100, so we will test them too.
We then find that
$L(0)=160$
$L(\sqrt{1800})=131.06$
$L(100)=116.62$
This means that the minimum possible length for cable is 116.62 feet.

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**Problem 2.** Use calculus to find the point $(x,y)$ on the parabola traced out by $y = x^2$ that is closest to the point $(3,0)$.
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<iframe src="https://www.desmos.com/calculator/5wltekcjy6?embed" width="500px" height="500px" style="border: 1px solid #ccc" frameborder=0></iframe>

We have the function $y=x^2$ and we want to minimize the the distance function $D=\sqrt{\left(x-3\right)^{2}+\left(y-0\right)^{2}}$
$D=\sqrt{\left(x-3\right)^{2}+\left(y\right)^{2}}$
$D=\sqrt{\left(x-3\right)^{2}+\left(x^2\right)^{2}}$
$D=\sqrt{x^{4}+x^{2}-6x+9}$
Now we find the derivative of this distance function.
$D'=\frac{1}{2}(x^4+x^2-6x+9)^{-\frac{1}{2}}*4x^3+2x-6$
$D'=\frac{2x^3+x-3}{\sqrt{x^{4}+x^{2}-6x+9}}$
The Function $D'$ is defined at every point. We then set $D'$ equal to zero.
$0=\frac{2x^3+x-3}{\sqrt{x^{4}+x^{2}-6x+9}}$
$0=2x^3+x-3$
$0=(2x+3)(x-1)$
$x=\frac{-3}{2}$ or $x=1$ are the critical points of the function which we will test to find the least distance.
As seen in the attached image, doing the first derivative test reveals that at $x=1$ the distance from $(3,0)$ to $y=x^2$ is minimized.
$(1,1)$ is the point on $y=x^2$ that is the closest to $(3,0)$.

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**Problem 3.** Use calculus for find the maximum possible area of a right triangle under the curve
$$
f\left(x\right)=x\left(x-4\right)^4
$$
in the first quadrant with one corner at the origin and one side along the $x$-axis.
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