Science:Math Exam Resources/Courses/MATH102/December 2019/Question 12/Solution 1

Denote the position where we cut the wire as $x$. We construct the triangle with length $x$ and the rectangle with length $90-x$.

Note that the perimeter of the triangle is $x$. Therefore, the length of its side is $x/3$. The area of the triangle is ${\frac {\sqrt {3}}{4}}\left({\frac {x}{3}}\right)^{2}$.

Similarly, let $\ell$ be length of the shorter side of the rectangle. Then, the perimeter is $2\ell +4\ell =6\ell =90-x$. The length is $\ell ={\frac {90-x}{6}}$ and its area is $2\left({\frac {90-x}{6}}\right)^{2}$.

The area to be maximized can be written as $A(x)={\frac {\sqrt {3}}{4}}\left({\frac {x}{3}}\right)^{2}+2\left({\frac {90-x}{6}}\right)^{2}$. We compare its area at the critical point and the end points.

Differentiating the area, ${\frac {dA}{dx}}={\frac {2{\sqrt {3}}}{36}}x-4\left({\frac {90-x}{36}}\right)$. Solving for the critical point gives us $2{\sqrt {3}}x-360+4x\implies x=360/(2{\sqrt {3}}+4)$.

However, note that this function attains its local minimum at the critical point. Checking values at the end points $A(0){\text{ and }}A(90)$, we can see that the total area is maximized when $x=0$, so when all the wire is used to create the rectangle.