Science:Math Exam Resources/Courses/MATH100/December 2012/Question 07/Solution 1

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 $\left({\frac {1}{2}},{\frac {\sqrt {3}}{2}}\right)$