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Chapter 3: Problem 22

A rectangle is bounded by the \(x\) - and \(y\) -axes and the graph of \(y=(6-x) / 2\) (see figure). What length and width should the rectangle have so that its area is a maximum?

Short Answer

Expert verified
The rectangle should have a length of 3 units and width of 1.5 units for its area to be a maximum of 4.5 square units.

Step by step solution

01

Formulate the Area Function

The area A of the rectangle is given by the product of its length and width. Here, one side is along the x-axis, let's call it 'x'; and the other side lies along the y-axis/f(x), let's call it 'y'. For our case, y is given by \((6-x)/2\). Therefore, the area function is \(A(x)=x(6-x)/2\).
02

Determine the Domain

By noting that the rectangle is bounded by x-axis and lies in the first quadrant, the domain of x is \(0 \leq x \leq 6\). Outside this interval, the area wouldn't make sense (negative area or area when the rectangle side is not on the line).
03

Find the Derivative of the Area Function

Find \(A'(x)\) and set it to zero to determine the critical points. \(A'(x) = 3-x\). Set \(A'(x)=0\) and solve for x, which gives \(x = 3\). This is within our domain so we take it as a valid critical point.
04

Find the Maximum Area

Test the endpoints and the critical point in the area function to determine the maximum area. These points are x=0, x=3, and x=6. Calculate the area at these points: \(A(0) = A(6) = 0\), but \(A(3) = (3*3)/2 = 4.5\). Thus, the maximum area of 4.5 square units is achieved when \(x = 3\). The length and width of the rectangle for maximum area thus are 3 units and \((6-3)/2 = 1.5\) units respectively.

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