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 two verticle poles with cables and their bases will make two right triangles. The Pythagora's therum can be used.
$A=\sqrt{80^{2}+x^{2}}=\sqrt{6400+x^{2}}$
$B=\sqrt{60^{2}+\left(100-x^{2}\right)}=\sqrt{13600+x^{2}-200x}$
The total length of the wire would equal:
$y'\left(x\right)=\frac{x}{\sqrt{6400+x^{2}}}+\frac{\left(x-100\right)}{\sqrt{13600+x^{2}-200x}}$
Now we have to find the critical points so we must solve:
$y'\left(x\right)=0$
$=\frac{x}{\sqrt{6400-x^{2}}}+\frac{\left(x-100\right)}{\sqrt{13600+x^{2}-200x}}=0$
$x=\frac{400}{7}$
Now we must find the double derivitive of y(x):
$y''\left(x\right)=\frac{x^{2}}{x^{2}+6400^{\frac{3}{2}}}+\frac{1}{\sqrt{6400+x^{2}}}+\frac{1}{\sqrt{x^{2}[phone]x}}-\frac{\left(2x-200\right)^{2}}{\left(x^{2}[phone]x\right)^{\frac{3}{2}}}$
Now we have to see if y''(x) at $x=400/7$ is positive or negative.
$y''\left(\frac{400}{7}\right)=\frac{301}{17760\sqrt{74}}>0$
Now as the second derivitive at $x=400/7$ is positive and there will be a minimum at $x=400/7$
Now $y(400/7)\sqrt{\left(6400+\frac{400}{7}\right)^{2}}+\sqrt{13600+\left(\frac{400}{7}\right)^{2}-200\left(\frac{400}{7}\right)}$
$y\left(\frac{400}{7}\right)=20\sqrt{74}=172.05$ feet, is the minimum length of cable required.

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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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$d^{2}=\left(x-3\right)^{2}+\left(y-0\right)^{2}$
$d^{2}=\left(x-3\right)^{2}+y^{2}$
$d^{2}=x^{2}-6x+y+x^{4}$
$d^{2}=x^{4}+x^{2}-6x+y$
$f\left(x\right)=x^{4}+x^{2}-6x+y$
$f'\left(x\right)=4x^{3}+2x-6=0$
$\left(x-1\right)=0$
$x=1$
$y=x^{2}=\left(1\right)^{2}=1$
The point is $(1,1)$

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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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