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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We want to minimize the possible length of the cable.

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By using the Pythagorean Theorem, we can minimize the length of the cable.

$a^2+b^2=c^2$
$c=\sqrt{60^2}+(100−𝑥)^2$
$=\sqrt{13600-200x+x^2}$

$a^2+b^2=c^2$
$c=\sqrt{80^2+x^2}$
$=\sqrt{6400+x^2}$

$C(x)=\sqrt{13600-200x+x^2}+\sqrt{6400+x^2}$
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The function found will now output the critical values by using derivative of the function. These critical values will be between 0 and 100.

$C'(x)=\frac{d}{dx}[(\sqrt{13600-200x+x^2})+(\sqrt{6400+x^2})]$

$C'(x)=[\frac{1}{2}(13600-200x+x^2)^\frac{-1}{2}*\frac{d}{dx}[13600-200x+x^2]]+[\frac{1}{2}(6400+x^2)^\frac{-1}{2}*\frac{d}{dx}[6400+x^2]]$

$C'(x)=[\frac{1}{2}(13600-200x+x^2)^\frac{-1}{2}*(-200+2x)]+[\frac{1}{2}(6400+x^2)^\frac{-1}{2}*(2x)]$

$C'(x)=\frac{-200+2x}{2\sqrt{13600-200x+x^2}}+\frac{x}{\sqrt{6400+x^2}}$
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C’ is defined everywhere, set the derivative to 0 to find the critical values,

$\frac{-200+2x}{2\sqrt{13600-200x+x^2}}+\frac{x}{\sqrt{6400+x^2}}=0$

$\sqrt{x^2+6400}(𝑥−100)+𝑥\sqrt{x^2−200𝑥+13600}=0$

$\sqrt{x^2+6400}(𝑥−100)=-𝑥\sqrt{x^2−200𝑥+13600}$

$(\sqrt{x^2+6400}(𝑥−100))^2=(-𝑥\sqrt{x^2−200𝑥+13600})^2$
$(x^2+6400)(𝑥−100)^2=(x^2−200𝑥+10000)(x^2+6400)$

$(x^2∗x^2+6400x^2−200x^2−200∗6400x+10000x^2+10000∗6400(x^4+16400x^2−200x^3−1280000x+64000000)=$
$(x^4−200x^3+13600x^2)$

$(x^4−16400x^2−200x^3−1280000x+64000000)−(x^4−200x^3+13600x^2)=$
$(x^4−200x^3+13600x^2)−(x^4−200x^3+13600x^2)$

Simplify

$2800x^2−1280000x+64000000=0$
$400(7x^2−3200𝑥+160000)$
$400(7x−400)(x−400)$

Set $=0$ to get critical values.

$7x-400=0$
$7x=400$
$x=\frac{400}{7}=57.14$
and
$x-400=0$
$x=400$
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Because 400 is greater than 100 and the cable cannot be greater than 100 due to the towers being 100ft apart, we will only use 400/7 critical value in our first derivative test.

$C'(x)=\frac{-200+2x}{2\sqrt{13600-200x+x^2}}+\frac{x}{\sqrt{6400+x^2}}$
$C'(2)=\frac{-200+2(2)}{2\sqrt{13600-200(2)+(2)^2}}+\frac{2}{\sqrt{6400+(2)^2}}$
$C'(2) <0$

$C'(x)=\frac{-200+2x}{2\sqrt{13600-200x+x^2}}+\frac{x}{\sqrt{6400+ x^2}}$
$C'(65)=\frac{-200+2(65)}{2\sqrt{13600-200x+(65)^2}}+\frac{65}{\sqrt{6400+(65)^2}}$
$C'(65)>0$


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$C(x)=\sqrt{13600-200x+x^2}+\sqrt{6400+x^2}$
$C(x)=\sqrt{13600-200(\frac{400}{7})+(\frac{400}{7})^2}+\sqrt{6400+(\frac{400}{7})^2}$
$=172.046ft$

The minimum possible length of the cable required is 172.046 ft.

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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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Since $y=x^2$ we can label the point (3,0) on the graph.

Using the Distance Formula we will find the distance between the point (3, 0) and the point $(x,y)$.
$d=\sqrt{( x_{2}−x_{1})^2 + (y_{2} − y_{1})^2}$
Since $𝑦=x^2$, we can substitute the y-value with $x^2$

$𝐷=\sqrt{(𝑥−3)^2+(x^2−0)^2}$
$𝐷=\sqrt{x^2−6x+9+x^4}$
$𝐷=\sqrt{x^4+x^2−6x+9}$
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Next, we find the derivative of D in order to find the critical values using the Chain Rule.

$D'=\frac{1}{2}(x^4+x^2-6x+9)^\frac{-1}{2}*(4x^3+2x-6)$
$D'=\frac{4x^3+2x-6}{2\sqrt{x^4+x^2-6x+9}}$
$D'=\frac{2x^3+x-3}{\sqrt{x^4+x^2-6x+9}}$
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Find the critical values, set the numerator $=0$

$2x^3+x-3=0$
$(x−1)(2x^2+2x−3)=0$

$x-1=0$
$x=1$

and

$(2x^2+2x−3)=0$
$\frac{-2±\sqrt{2^2-4(2)(-3)}}{2(2)}$
$\frac{-2±\sqrt{4-24}}{4}$
$\frac{-2±\sqrt{-20}}{4}$
$=$ no solution, unable to take the square root of a negative number.
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1st Derivative Test


The first derivative test shows that there is a minimum at $f(1)$, getting the point $(1,f(1))$ which also gives us the final point of $(1,1)$.

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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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