Science:Math Exam Resources/Courses/MATH110/April 2014/Question 05/Solution 1

We can proceed by using the standard template for solving optimisation questions. The exact steps might vary between course and section, but the overall steps will be similar.

1. Let's define the coordinates of $P=(x,y)$.

2. The area of the triangle is therefore: $A={\frac {1}{2}}(1+x){\sqrt {1-x^{2}}}$

3. The area equation is already in one variable. The domain is $x\in [-1,1]$. In that we can have $x$ between $-1$ and $1$ (the diameter of the circle) and we can include the endpoints since this will give an area of $0$.

4. Differentiate:

${\begin{aligned}A'(x)&={\frac {1}{2}}\left[(1+x)\cdot {\frac {-2x}{2{\sqrt {1-x^{2}}}}}+1\cdot {\sqrt {1-x^{2}}}\right]\\&={\frac {1}{2}}\left[{\frac {-x-x^{2}+(1-x^{2})}{\sqrt {1-x^{2}}}}\right]\\&={\frac {1}{2}}\left[{\frac {1-x-2x^{2}}{\sqrt {1-x^{2}}}}\right]\end{aligned}}$

In this case, the critical points are when the numerator is equal to zero (denominator is zero when $x=\pm 1$).

${\begin{aligned}x&={\frac {1\pm {\sqrt {1-4(-2)(1)}}}{-4}}\\&={\frac {1\pm {\sqrt {9}}}{-4}}={\frac {1}{2}},-1\end{aligned}}$

We can then use the closed interval method:

${\begin{aligned}A(1)&=0\\A(-1)&=0\\A\left({\frac {1}{2}}\right)&={\frac {1}{2}}\left(1+{\frac {1}{2}}\right){\sqrt {1-\left({\frac {1}{2}}\right)^{2}}}={\frac {3{\sqrt {3}}}{8}}>0\end{aligned}}$

5. The maximum area happens at point $P=\left({\frac {1}{2}},{\frac {\sqrt {3}}{2}}\right)$.