Science:Math Exam Resources/Courses/MATH100/December 2012/Question 07

MATH100 December 2012

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Question 07

Full-Solution Problems. In questions 5-11, justify your answers and show all your work. If a box is provided, write your final answer there. Unless otherwise indicated, simplification of numerical answers is required in these questions. Find the coordinates of the point P, which lies on the curve $y={\sqrt {1-x^{2}}}$, in the diagram below for which the area of the right triangle ABP is the maximum. Remember to justify that your answer actually gives the maximum area.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
What is the base of the triangle in terms of x? The height? Then express the area of the triangle in terms of x and find the maximum of this function.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We are trying to maximize the area of triangle ABP, which is given by the formula: $A={\frac {1}{2}}bh$ As shown in the picture above, the base of triangle ABP is $(1+x)$ and the height is the y-value of point P, or ${\sqrt {1-x^{2}}}$, so our final formula for area is: $A(x)={\frac {1}{2}}(1+x){\sqrt {1-x^{2}}}$ where $-1\leq x\leq 1$, as x cannot extend past the radius of the circle, which is 1. The maximum of this function will occur either at the endpoints of the domain or critical points in the open interval $(-1,1)$. Thus, the endpoints $x=-1$, $x=1$ will be two of the points we test to find the maximum To find remaining critical points in the interval $(-1,1)$, we take the derivative and find where it is either equal to zero or undefined. Using the product rule, we get: $A'(x)={\frac {1}{2}}\left({\sqrt {1-x^{2}}}+{\frac {(1+x)(-2x)}{2{\sqrt {1-x^{2}}}}}\right)$ Finding a common denominator, and combining terms on the top gives: ${\begin{aligned}A'(x)&={\frac {1}{2}}\left({\frac {2-2x^{2}-2x^{2}-2x}{2{\sqrt {1-x^{2}}}}}\right)\\&={\frac {1}{2}}\left({\frac {2(1-x-2x^{2})}{2{\sqrt {1-x^{2}}}}}\right)\\&={\frac {1}{2}}\left({\frac {1-x-2x^{2}}{\sqrt {1-x^{2}}}}\right)\end{aligned}}$ Setting equal to zero, we get: ${\begin{aligned}0&={\frac {1}{2}}\left({\frac {1-x-2x^{2}}{\sqrt {1-x^{2}}}}\right)\\&={\frac {1-x-2x^{2}}{\sqrt {1-x^{2}}}}\\&={\frac {(1+x)(-2x+1)}{\sqrt {1-x^{2}}}}\end{aligned}}$ On the top, $x=-1$ and $x=1/2$ gives us zero, while on the bottom, $x=1$ and $x=-1$ makes the derivative undefined. Thus, we test the endpoints, $x=-1,x=1$, and the critical point $x=1/2$ in the area formula $A(x)$ to determine which gives the maximum area. ${\begin{array}{llll}A(-1)&={\frac {1}{2}}(1+(-1)){\sqrt {1-(-1)^{2}}}&={\frac {1}{2}}(0)(0)&=0\\A(1)&={\frac {1}{2}}(1+(1)){\sqrt {1-(1)^{2}}}&={\frac {1}{2}}(2)(0)&=0\\A(1/2)&={\frac {1}{2}}(1+(1/2)){\sqrt {1-(1/2)^{2}}}&={\frac {1}{2}}\left({\frac {3}{2}}\right){\sqrt {\frac {3}{4}}}&>0\end{array}}$ So the area is maximized when $x=1/2$. The associated point P is given by $\left({\frac {1}{2}},{\sqrt {1-\left({\frac {1}{2}}\right)^{2}}}\right)$ or $\left({\frac {1}{2}},{\frac {\sqrt {3}}{2}}\right)$

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