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Chapter 4: Problem 24

A rectangle is constructed with its base on the \(x\) -axis and two of its vertices on the parabola \(y=16-x^{2} .\) What are the dimensions of the rectangle with the maximum area? What is that area?

Short Answer

Expert verified
Answer: The dimensions of the rectangle with the maximum area are length (base) = \(4\sqrt{2}\) and height = 8.

Step by step solution

01

Define the variables needed for the area formula

Let \(x\) be the distance from the y-axis; therefore, the x-coordinates of the vertices will be \(x\) and \(-x\). Then the base of the rectangle will have a length of \(2x\). The y-coordinate of these vertices can be found using the parabola equation \(y = 16 - x^2\).
02

Find the formula of the area of the rectangle

The area of the rectangle can be determined by multiplying its base by its height. We know that the base of the rectangle is \(2x\) and that the height is given by \(16-x^2\). So the area can be given by the following formula: $$ A(x) = (2x)(16-x^2) = 32x - 2x^3 $$
03

Find the critical points of the area function

We need to find the maximum area, so we need to differentiate the area function and set it to zero: $$A'(x) = \frac{d}{dx}(32x - 2x^3) = 32 - 6x^2$$ Set the derivative equal to zero and solve for x: $$32 - 6x^2 = 0 \Rightarrow x^2 = \frac{32}{6} \Rightarrow x = \pm \sqrt{\frac{32}{6}} = \pm 2\sqrt{2}$$
04

Choose the relevant critical point

Since we are dealing with the \(x\) -axis, we know that a negative value of \(x\) is not relevant for our problem. Therefore, we choose the positive critical point: $$x = 2\sqrt{2}$$
05

Find the dimensions of the rectangle

Now, find the height of the rectangle by plugging the x-value into the parabola equation: $$y = 16 - x^2 = 16 - (2\sqrt{2})^2 = 16 - 8 = 8$$ So, the dimensions of the rectangle are: Length (base): \(2x = 2(2\sqrt{2}) = 4\sqrt{2}\) Height: \(y = 8\)
06

Compute the maximum area

Using the area formula and the dimensions of the rectangle, calculate the maximum area: $$ A(x) = (4\sqrt{2})(8) = 32\sqrt{2} $$ Therefore, the maximum area of the rectangle is \(32\sqrt{2}\).

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