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 know both triangles are right triangles thus we use,
Formula: Pythagoream Theorem
A=$\sqrt{\left(80\right)^{2}+\left(100-x\right)^{2}}$
B=$\sqrt{\left(60\right)^{2}+x^{2}}$
We obtain total length of cable wire by:$
L=A+B=$\sqrt{6400+\left(100-x\right)^{2}}$+$\sqrt{x^{2}+3600}$
But We want to obtain minimum cable length thus,
$d(L=A+B)/dx$=0
L'=$\left(\frac{1}{2}\left(x^{2}+3600\right)^{-\frac{1}{2}}\cdot\frac{d}{dt}\left(x^{2}+3600\right)\right)+\left(\frac{1}{2}\left(6400+x^{2}\right)-200x+10,000\right)^{-\frac{1}{2}}\cdot\frac{d}{dt}\left(6400+x^{2}-200x+10,000\right)$
L'=$\left(\frac{1}{2}\left(x^{2}+3600\right)^{-\frac{1}{2}}\cdot\left(2x\right)\right)\left(\frac{1}{2}\left(6400+x^{2}-200x+10,000\right)^{-\frac{1}{2}}\cdot\left(2x-200\right)\right)$
L'=$\frac{x}{\sqrt{x^{2}+3600}}+\frac{\left(x-100\right)}{\sqrt{x^{2}-200x+16400}}$
Set L'=0 and Solve:
L'=$\left(\frac{1}{2}\left(x^{2}+3600\right)^{-\frac{1}{2}}\cdot\left(2x\right)\right)\left(\frac{1}{2}\left(6400+x^{2}-200x+10,000\right)^{-\frac{1}{2}}\cdot\left(2x-200\right)\right)$
$\frac{x}{\sqrt{x^{2}+3600}}+\frac{\left(x-100\right)}{\sqrt{x^{2}-200x+16400}}=0$
$\sqrt{\frac{x}{\sqrt{x^{2}+3600}}}\ =\sqrt{-\frac{\left(x-100\right)}{\sqrt{x^{2}-200x+16400}}}$
$\frac{x^{2}}{x^{2}+3600}\ =\frac{\left(x-100\right)^{2}}{x^{2}-200x+16400}$
Crossed Multiplied both fractions:
$\left(7x-300\right)\left(x+300\right)=0$
Thus my Critical Values are:
x=-300 and x=300/7
Only use positive x-values $x=300/7$

$1st$ D $Test:$

L has a minimum at 300/7
Substitution for $L(300/7)=
$L\left(\frac{300}{7}\right)=\sqrt{\left(\frac{300}{7}\right)^{2}+3600}+\sqrt{16400+\left(100-
\frac{300}{7}\right)^{2}}$
$=\frac{60}{7}\sqrt{74}+\frac{80}{7}\sqrt{74}=\frac{140}{7}\sqrt{74}$
$=20\sqrt{74\ }=172ft$
Conclusion:
Minimum possible length of cable is 172ft.

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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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Distance Fromula:
D=$\sqrt{\left(x-3\right)^{2}+\left(y^{2}\right)}$
Substitute in $y=x^2$
D=$\sqrt{\left(x-3\right)^{2}+x^{4}}$
D=$\sqrt{x^{2}-6x+9+x^{4}}$

Derivative
D'=$\frac{1}{2}\left(x^{4} +x^{2} +9\right)^{-\frac{1}{2}}\cdot\left(4x^{3}-+2x-6\right)$
D'=$\frac{\left(4x^{3}+2x^{2}-6\right)}{2\sqrt{x^{4}+x^{2}-6x+9}}$
Note D' is defined everywhere
Set D'=0 and solve:
(Note when setting a fraction=0 the numberator is equal to zero.)
$0=2x^3+x-3$
$x=1$ critical value
1st Derivative Test:

Conclusion:
D is minimize at $x=1$
to figure which is the actual minimal value we have to plug them into
$y=x^2$
$y=(1)^2$
The closest point on the graph of $y=x^2$ to point (3,0) 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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