A rectangular poster is to contain 242 square inches of print. The margins at the top and bottom of the poster are to be 2 inches, and the margins on the left and right are to be 1 inch. What should the dimensions of the poster be (in inches) so that the least amount of poster is used?

To solve this problem, we need to use calculus.

Let's denote the width of the printed area as x (in inches) and the height as y (in inches).

The area of the printed area is given by the equation: Area = x*y = 242 square inches.

The total area of the poster, including the margins, is (x+2)*(y+4) because there are 1 inch margins on either side of the width (adding 2 inches to the total width) and 2 inch margins on the top and bottom of the height (adding 4 inches to the total height).

We want to minimize the total area of the poster, so we need to find the minimum of the function (x+2)*(y+4).

We can express y in terms of x using the equation for the area of the printed area: y = 242/x.

Substitute y = 242/x into the equation for the total area of the poster to get a function of x: Total Area = (x+2)*((242/x)+4).

Now we can find the derivative of this function with respect to x and set it equal to zero to find the value of x that minimizes the total area of the poster.

The derivative of the function is: d(Total Area)/dx = 4 - 484/x^2.

Setting this equal to zero gives: 4 - 484/x^2 = 0.

Solving for x gives: x = sqrt(484/4) = 11 inches.

Substitute x = 11 inches into the equation y = 242/x to find y: y = 242/11 = 22 inches.

So, the dimensions of the poster that minimize the total area are 11 inches (width of the printed area) + 2 inches (margins) = 13 inches and 22 inches (height of the printed area) + 4 inches (margins) = 26 inches.