Q. 7 - Precalculus Enhanced with Graphing Utilities

Question

A rectangle has one corner in quadrant I on the graph of $y=16-x^2$, another at the origin, a third on the positive y-axis and the fourth on the positive x-axis. See the figure.

Part (a): Express the area A of the rectangle as a function of x.

Part (b): What is the domain of A?

Part (c): Graph $A=A\left(x\right)$. For what value of x is A largest?

Step-by-Step Solution

Answer

Part (a): The area A of the rectangle as a function is $A\left(x\right)=x\left(16-x^2\right)$.

Part (b): The domain of A is $\left(0,4\right)$.

Part (c): On plotting the function, we get,

The value of x  for which A is the largest is $x\approx 2.31$.

Part (a) Step 1. Given information.

Consider the given figure,

Distance between the points $\left(0,0\right),\left(x,0\right)$ is the length of the rectangle,

$$\begin{gathered} l=\sqrt{{\left(x-0\right)}^2+{\left(0-0\right)}^2} \\ =\sqrt{x^2} \\ =x \end{gathered}$$

Part (a) Step 2. Find the breadth.

Consider the given figure,

Distance between the points $\left(x,0\right),\left(x,y\right)$ is the breadth of the rectangle,

$$\begin{gathered} b=\sqrt{{\left(x-x\right)}^2+{\left(y-0\right)}^2} \\ =\sqrt{y^2} \\ =y \\ =16-x^2 \end{gathered}$$

We know the area of the rectangle is $A=b\times l$.

Substitute the values,

$A\left(x\right)=x\left(16-x^2\right)$

Part (b) Step 1. Given information.

Consider the given question,

Area of the given rectangle is $x\left(16-x^2\right)$.

As $A>0$, then $x>0$ and $x<4$.

Therefore, the domain is $\left(0,4\right)$.

Part (c) Step 1. Plot the function.

On plotting the function, we get,

From the graph, we can say that the largest value of x is largest A at $x\approx 2.31$