Problem 27 - Essential Calculus

Question

A rectangular page is to contain 30 square inches of print. The margins on each side are 1 inch. Find the dimensions of the page such that the least amount of paper is used.

Step-by-Step Solution

Answer

If the printed area measures \(\sqrt{15}\) inches in width and \(2\sqrt{15}\) inches in length, then the dimensions of the page that minimizes the amount of paper used are \(\sqrt{15} + 2\) inches wide and \(2\sqrt{15} + 2\) inches long.

Step 1: Setup

Let the width of the printed area be \(x\) inches, and the length be \(y\) inches. Therefore, we have: \(xy=30\) (given that the printed area is 30 square inches).

Step 2: Formulate the area function

The total dimensions of the page are \((x+2)\) inches wide and \((y+2)\) inches long (considering the 1 inch margin on all sides). Hence, the total area 'A' of the page can be given by: \(A=(x+2)(y+2)\).

Step 3: Express the Area as a function of a single variable

From our first equation, we can express \(y\) as \(y=30/x\). Substituting this into our area equation gives us: \(A=(x+2)(30/x+2)\).

Step 4: Differentiation

To find the minimum of 'A', we can take the derivative and set it to zero. It simplifies to: \(A' = 30/x^2 - 4/x\).

Step 5: Finding Minimum

Setting \(A'\) to zero and solving for \(x\), gives us: \(x^2=15\). Therefore, \(x=\sqrt{15}\). Substitute \(\sqrt{15}\) into our equation for \(y\) to get: \(y=30/\sqrt{15}=2\sqrt{15}\). Giving us the dimensions of the page that minimizes the paper used.

Step 6: Check solution

Taking second derivative of A and substituting \(x=\sqrt{15}\) yields a positive value confirming that this indeed gives a minimum. Therefore, the dimensions that minimizes the area of paper used are width \((x+2)\) and length \((y+2)\).