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title: 'Math 181 Miniproject 8: Applied Optimization'

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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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**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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The goal is to minimize the distance between the two points:$D=\sqrt{\left(x-3\right)^{2}+\left(y-0\right)^{2}}$
$D=\sqrt{x^{4}+x^{2}-6x+9}$
$D'=\frac{2x^{3}+x-3}{\sqrt{x^{4}+x^{2}-6x+9}}$
I then set $D'$ equal to zero to find the minimum.
$0=\frac{2x^{3}+x-3}{\sqrt{x^{4}+x^{2}-6x+9}}$
$x=1$
After conducting the 1st-Derivative test, $x=1$ is confirmed as the minimum.
The point on the parabola closest to $(3,0)$ is (1,1).
![](https://i.imgur.com/FRMgxXE.jpg)



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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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I want to maximize the area of the right triangle:
$A=\frac{1}{2}bh$
$A=\frac{1}{2}xy$
$A=\frac{1}{2}x\left(x\left(x-4\right)^{4}\right)$
$A=\frac{1}{2}x^{6}-8x^{5}+48x^{4}-128x^{3}+128x^{2}$
$A'=3x^{5}-40x^{4}+192x^{3}-384x^{2}+256x$
I then set $A'$ equal to zero to find the maximum.
$0=3x^{5}-40x^{4}+192x^{3}-384x^{2}+256x$
$x=0,\frac{4}{3},4$
After conducting the 1st-Derivative test, $x=\frac{4}{3}$ is confirmed as the maximum value of $x$ (base).
The hieght of the triangle is $f\left(x\right)=\frac{4}{3}\left(\frac{4}{3}-4\right)^{4}=67.424$
The maximum area of the traingle under the curve $f\left(x\right)=x\left(x-4\right)^4$ is $A=44.949$
![](https://i.imgur.com/zyixIYq.png)



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