Chapter 4.6, Problem 18E

To determine

To calculate: The dimension of the rectangle of largest area that has its base on the $x-\text{axis}$ and its other two vertices above the $x-\text{axis}$ and lying on the parabola $y=8-x^2$

Answer to Problem 18E

The dimensions of the rectangle of largest area that has its base on the $x-\text{axis}$ are $\sqrt{\frac{8}{3}}\text{ and }\frac{64\sqrt{2}}{3\sqrt{3}}$

Explanation of Solution

Given information:

The rectangle of largest area that that has its base on the $x-\text{axis}$ and its other two vertices above the $x-\text{axis}$ and lying on the parabola $y=8-x^2$

Formula used:

Let $2x$ be the length and $y$ be the breadth of rectangle.

Area of rectangle is $A=2x\times y$

And

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at $f^{\prime }(x)>0$ every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3.   $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$is neither a point of local maxima nor a point of local minima.

Calculation:

As per the given problem

A rectangle of largest area that has its base on the $x-\text{axis}$ and its other two vertices lying on the parabola

  $y=8-x^2\mathrm{......}\left(1\right)$

Recall that, Let $2x$ be the base length and $y$ be the breadth of rectangle.

Area of rectangle is

  $A=2x\times y$

Substitute $y=8-x^2$ , to get

  $\begin{array}{l}A\left(x\right)=2x\times \left(8-x^2\right)\\ A\left(x\right)=16x-2x^3\end{array}$

Recall that,

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

Differentiate on both sides,

  $A^{\prime }\left(x\right)=16-6x^2$

Solve for $A^{\prime }\left(x\right)=0$ , and simplified

  $\begin{array}{l}16-6x^2=0\\ -6x^2=-16\\ x^2=\frac{16}{6}\\ x^2=\frac{8}{3}\end{array}$

Take square root on both sides, to get

  $x=\pm \sqrt{\frac{8}{3}}$

The equation has two real solutions but length can’t be negative,

Therefore,

  $x=\sqrt{\frac{8}{3}}$

Substitute $x=\sqrt{\frac{8}{3}}$ in equation $(1)$ and simplified

  $\begin{array}{l}y=8-x^2\\ =8-{\left(\sqrt{\frac{8}{3}}\right)}^2\\ =8-\frac{8}{3}\\ =\frac{24-8}{3}\\ =\frac{16}{3}\end{array}$

Therefore,

The maximum area of the rectangle is

  $\begin{array}{l}A=2\times \sqrt{\frac{8}{3}}\times \frac{16}{3}\\ =32\times 2\frac{\sqrt{2}}{3\sqrt{3}}\\ =\frac{64\sqrt{2}}{3\sqrt{3}}\end{array}$

The dimensions of the rectangle of largest area that has its base on the $x-\text{axis}$ are $\sqrt{\frac{8}{3}}\text{ and }\frac{64\sqrt{2}}{3\sqrt{3}}$