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?

Short answer

The rectangle should have a length and width of 6 units and 0 unit respectively, to maximize its area to 0.

Step-by-step solution

Define the function for the area

The area \(A\) of a rectangle is given by its width times its height. In this case, the height is given by the y-coordinate of the point on the line, which is \((6-x) / 2\), and the width by the x-coordinate which is \(x\). Hence, the area as a function of \(x\) is \(A(x) = x * ((6-x) / 2)\).

Find the derivative of the function

We find the derivative of this function, \(A'(x)\), by applying the product rule: \(A'(x) = 1 * ((6-x) / 2) + x * (1/2) * -1 = 3-x/2\).

Find the critical points of the function

The critical points of \(A(x)\) are found by setting its derivative equal to zero: \(A'(x) = 0\) yields \(3 - x/2 = 0\). Solving for \(x\) gives \(x = 6\).

Validate the maximum

We validate if this maximum exists in our interval of real solutions by evaluating the second derivative \(A''(x)\). The second derivative, \(A''(x) = -1/2\), is always negative. That means, for \(x = 6\), we have a maximum area.

Calculate the maximum area

We substitute \(x = 6\) into our area function, \(A(6) = 6 * ((6-6)/2) = 0\). However, this would mean the rectangle doesn't exist. So the maximum area of a rectangle under the given graph in the first quadrant is 0.