Math 181 Miniproject 8: Applied Optimization.md
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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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For this problem, we want to minimize distance of L1+L2= L3

$D=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

L1 = $\sqrt{60^{2}+x^{2}}$
L1 = $\sqrt{60^{2}+x^{2}}$ --- Next, take the derivative of L1

L1' = $\left(\frac{1}{2}\left(3600+x^{2}\right)^{\left(-\frac{1}{2}\right)}\right)\left(2x\right)$

L1'= $\frac{\left(2x\right)}{2\sqrt{3600+x^{2}}}$

Next, I did this same process to the other side to find the distance of the cable of the 80 ft tower.

$L_{2}=\sqrt{80^{2}+\left(100-x\right)^{2}}$

$L_{2}=\sqrt{16400-200x+x^{2}}$

$L_{2}'=\frac{1}{2}\left(16400-200x+x^{2}\right)^{\left(-\frac{1}{2}\right)}\left(2x-200\right)$

$L_{2}'=\frac{\left(2x-200\right)}{2\sqrt{16400+x^{2}-200x}}$

Now that we found the L'1 and L'2, we can set the derivatives of L3 equal to 0 to find our critical number.

$L_{3}=\ \sqrt{3600+x^{2}}+\sqrt{80^{2}+\left(100-x\right)^{2}}$

$L'3=\ \left(\frac{2x}{2\sqrt{3600+x^{2}}}\right)+\frac{\left(2x-200\right)}{2\sqrt{16400+x^{2}-200x}}$

$0=\ \left(\frac{2x}{2\sqrt{3600+x^{2}}}\right)+\frac{\left(2x-200\right)}{2\sqrt{16400+x^{2}-200x}}$

I used desmos to find citical numbers and found

$x= 42.857$ We can use this value to run a first derivative test to see if this gives us a minimum value.

$f'\left(0\right)<0$
$f'\left(100\right)>0$

With the first derivative test, we can say that x= 42.857 is a minimum value so we can ot plug this back into our orginal formula for L3

$L3=\sqrt{3600+\left(42.875\right)^{2}}+\sqrt{80^{2}+\left(100-42.875\right)^{2}}$

$L3= 172.04ft$

We cn say that the minimum amout of cable between the two towers is 172.04ft long.

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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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For this problem, we want to minimize distance using the formula:

$D=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}+y_{1}\right)^{2}}$

$D=\sqrt{\left(x-3\right)^{2}+y^{2}}$

$D=\ \sqrt{\left(x^{4}+x^{2}-6x+9\right)}$ --- Next, we find the derivative of D.

$D'=\ \frac{1}{2}\left(x^{4}+x^{2}-6x\right)^{\left(-\frac{1}{2}\right)}\left(4x^{3}+2x-6\right)$

$D'=\ \frac{\left(4x^{3}+2x-6\right)}{2\sqrt{x^{4}+x^{2}-6x+9}}$

From this point, we can cancel a 2 from the denominator and numerator to get:

$D'=\ \frac{\left(2x^{3}+x-3\right)}{\sqrt{x^{4}+x^{2}-6x+9}}$ We can set the numerator = 0 to get our critial value.

$0=2x^{3}+x-3$

$x=1$
Next, we run a first derivative test to see if x=1 is a minimum value. I decided to test for D’(0) and D’(2).

$D'\left(0\right)<0$
$D'\left(2\right)>0$

From this test, I am able to say that the distance is minimized when x=1. Lastly, I plugged in my value for x into the original formula $y=x^2$ to solve for y.

$y=(1)^2$
$y=1$

We can say that the closest point on the graph of $y=x^2$ to the point (3,0) is the point (1,1).

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