Question

A rectangular page is to contain 36 square inches of print. The margins at the top and bottom and on each side are to be \(1 \frac{1}{2}\) inches. Find the dimensions of the page that will minimize the amount of paper used.

Short answer

The dimensions that will minimize the amount of paper used are 9 inches by 12 inches.

Step-by-step solution

Define the Dimensions

Let's assign variables to the dimensions of the page. Denote the width (horizontal dimension) as \(x\) and the height (vertical dimension) as \(y\). The area of the print itself is 36 square inches. Because the print area is a rectangle itself and it's area is length x width, denote one dimension as \(x - 3\) (subtracting the side margins) and the other as \(\frac{36}{x - 3}\) (so that the area of the print remains constant at 36 square inches).

Formulate the Area of the Page

The area of the entire page, which includes the print and the margins, would be \(x*y\). Substituting our value for \(y\), which is the height of the print plus twice the top and bottom margin, we have \(Area = x * (2 * 1.5 + \frac{36}{x - 3}) = x * (\frac{36}{x - 3} + 3)\).

Find the Derivative

We need to find the derivative of the above function. Using the quotient rule and simplification, the derivative \(A'(x)\) is \(A'(x) = 3 - \frac{108}{(x - 3)^2}\).

Find the Zero of the Derivative

Set the derivative to zero and solve for \(x\). This will give us the value of \(x\) that minimizes the area. Solving the equation \(3 - \frac{108}{(x - 3)^2} = 0\), we get \(x = 9\).

Find the Corresponding y-value

Substitute \(x = 9\) into the formula for \(y\) to find the corresponding \(y\) value. This gives us \(y = 2 * 1.5 + \frac{36}{9 - 3} = 6 + \frac{36}{6} = 6 + 6 = 12\).

Verify the Minimum

Verifying this is indeed the minimum by the second derivative test. The second derivative is \(A''(x) = \frac{216}{(x - 3)^3}\) is positive when \(x = 9\). Hence, by the second derivative test, these dimensions minimize the area.