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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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We want to minimize. However, by drawing out the vertical poles with a connecting base of 100 ft apart. It is possible to have this form a bigger triangle. This allows to have a base of 100 ft but another side of 140 ft (60 ft +80 ft). By calculating this triangle, we can use the Pythagorean's theorem to find c. 
$a^{2}+b^{2}=c^{2}$
$140^{2}+100^{2}=c^{2}$
$19600+10000=c^{2}$
$29600=c^{2}$
$172.05=c$


![](https://i.imgur.com/EFvDBMi.jpg)




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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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We want to minimize. Since we know the $y=x^2$ is a parabola and want to find the closest point. We can use the distance forumla. 
$d=\sqrt{\left(x-0\right)^{2}+\left(y-0\right)^{2}}$
We know the point is closet to is (3,0). Therefore, we can plug this value in the distance formula.
$d=\sqrt{\left(x-3\right)^{2}+\left(y-0\right)^{2}}$
We could solve this, however, there are two variables in the formula. Thus, a variable must be substituted in the formula. In which, given was $y=x^2$, so we can use that information to be substitued in the distance formula. 
$d=\sqrt{\left(x-3\right)^{2}+\left(\left(x^{2}\right)-0\right)^{2}}$
We now can solve this.
$d=\sqrt{\left(x-3\right)\left(x-3\right)+x^{4}}$
$d=\sqrt{x^{2}-3x-3x+9+x^{4}}$
$d=\sqrt{x^{4}+x^{2}-6x+9}$
A derivative of this formula can now be found. 
$d'=\frac{1}{2}\left(x^{4}+x^{2}-6x+9\right)^{-\frac{1}{2}}\cdot\frac{d}{dx}\left[x^{4}+x^{2}-6x+9\right]$
$d'=\frac{1}{2}\left(x^{4}+x^{2}-6x+9\right)^{-\frac{1}{2}}\cdot\left(4x^{3}+2x-6\right)$
$d'=\frac{\left(2x^{3}+x-3\right)}{\sqrt{x^{4}+x^{2}-6x+9}}$
f' is defined everywhere, set f'=0
$2x^{3}+x-3=0$ to solve for x.
$x=1$
We should then test with the first derivative test. 


| Test Variable | Behavior of f' | Sign of f |
| -------- | -------- | -------- |
| f'(0)     | Negative   | Decreasing     |
| f'(100) | Positive | Increasing|

Analyzing the first derivative test, we see how it decreases and then increases. In other words, it provides evidence that there is minimum at this tested value of x=1.

Now that we tested that x=1, we can now use this back into the given formula for the parabola at $y=x^2$. 
$y=1^2$
$y=1$
Therefore, the point (1,1) on the parabola $y=x^2$ is closest to the point (3,0).

![](https://i.imgur.com/ZU7iOgm.png)



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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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