Science:Math Exam Resources/Courses/MATH110/April 2018/Question 09/Solution 1

By drawing the graph of the parabola and the rectangle,

we can see that the length of the the base is $2x$ and that of the height is $y=9-x^{2}$.

The base is positive if $x>0$ and the height is positive when $-3<x<3$. (See the graph of the parabola.) Therefore, we consider $x$ on $(0,3)$.

Since the area of a rectangle is the length of its base times that of its height, the area function $A$ can be defined as

$$A(x)=2x\cdot (9-x^{2}).$$

To find the maximum of $A$ on the interval $(0,3)$, we find the critical points first.

By the product rule for a derivative, the derivative of $A$ is

$$A'(x)=2(9-x^{2})+2x\cdot (-2x)=18-2x^{2}-4x^{2}=18-6x^{2}=6(3-x^{2})=6({\sqrt {3}}-x)({\sqrt {3}}+x).$$

Then, solving $A'(x)=0$, we get

$$x=-{\sqrt {3}},{\sqrt {3}}.$$

Since $-{\sqrt {3}}$ is outside of the interval $(0,3)$, we only consider ${\sqrt {3}}$. Observing that $A'(x)=6({\sqrt {3}}-x)({\sqrt {3}}+x)>0$ on $(0,{\sqrt {3}})$, $A$ is increasing on $(0,{\sqrt {3}})$. On the other hand, since $A'(x)=6({\sqrt {3}}-x)({\sqrt {3}}+x)<0$ on $({\sqrt {3}},3)$, $A$ is decreasing on $({\sqrt {3}},3)$. Therefore, we obtain the maximum value of the area $A$ at $x={\sqrt {3}}$.

Calculating the length of the base and height at $x={\sqrt {3}}$, we have

${\text{(the length of the base)}}=2{\sqrt {3}}$ and ${\text{(the length of the height)}}=9-({\sqrt {3}})^{2}=6$.

Answer: The length of the base and height are $\color {blue}2{\sqrt {3}},6$.