Problem 22 - Calculus of a Single Variable

Question

Maximum Area 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 dimensions of the rectangle that would yield the maximum area would be a width of 3 and a length of 1.5.

Step 1: Expression for the rectangle area

Let's denote the \(x\)-coordinate of the top right corner of the rectangle as \(x\), and the \(y\)-coordinate as \(y = (6-x)/2\). Thus, our rectangle has width \(x\) and length \((6-x)/2\), so its area can be expressed as \(A = x(6-x)/2\).

Step 2: Derive the area function

To find the maximum of the area \(A\), the derivative of the area function with respect to \(x\) is taken and set equal to zero. The derivative of \(A\) is given by \(A' = (6-x)-x\). Setting this equal to zero gives \(6-2x=0\).

Step 3: Compute the critical point

Solving for \(x\) from the equation obtained in the previous step gives \(x=3\). This is the \(x\)-coordinate of the rectangle's top right corner that will give a maximum area.

Step 4: Determine the dimensions

Inserting \(x=3\) into the equation for \(y\) gives the corresponding \(y\)-coordinate as \(y = (6-3)/2 = 1.5\). Therefore, the rectangle with the maximum area has dimensions: width = \(3\) and length = \(1.5\).