Problem 22 - Essential Calculus

Question

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?

Step-by-Step Solution

Answer

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 1: 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 \(A(x)=x(6-x)/2\).

Step 2: 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).

Step 3: 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.

Step 4: 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.