You are designing a rectangular poster to contain 256 in^2 of printing with a 3-in. margin at the top and bottom and a 2-in. margin at each side. What overall dimensions will minimize the amount of paper used?

let the width of the paper be x in

let the height of the paper be y in

then the width of the printed area is x-4 and the height of the printed area is y-5

we know (x-4)(y-5) = 256
xy - 5x - 4y + 20 = 256
y(x-4) = 5x + 236

y = (5x+236)/(x-4)

Area of page
= xy
= x(5x+236)/(x-4)
= (5x^2 + 236x)/(x-4)

Now differentiate that using the quotient rule, set it equal to zero and solve for x
note that x>4 to make sense, sub your x value back into the above y = ... relation to get the dimensions.

To minimize the amount of paper used, we want to find the dimensions of the rectangular poster that will contain 256 square inches of printing, while minimizing the area of the paper needed.

Let's assume the length of the rectangle is x inches and the width is y inches.

Given that there is a 3-inch margin at the top and bottom, and a 2-inch margin on each side, the overall dimensions of the poster can be expressed as:

Length = x + 2(2) = x + 4
Width = y + 2(3) = y + 6

The printing area can be represented as the product of the length and width:

Printing area = x * y

Given that the printing area is 256 square inches, we have:

x * y = 256

To minimize the amount of paper used, we need to minimize the total area of the rectangle, including the margins. The total area of the rectangle can be expressed as:

Total area = Length * Width = (x + 4)(y + 6)

Our goal is to minimize this total area while satisfying the condition x * y = 256.

To find the dimensions that minimize the total area, we can use the method of calculus.

1. Express the total area equation in terms of a single variable. Let's express it as a function of one variable, say x:
A(x) = (x + 4)(y + 6) = (x + 4)(256/x)

2. Differentiate the function A(x) with respect to x.
A'(x) = (0)(y + 6) + (1) * (256/x) - (256/x^2)(x + 4)

3. Set the derivative equal to zero and solve for x to find the possible value(s) of x that correspond to minimum area:
A'(x) = 0
(256/x) - (256/x^2)(x + 4) = 0
256 - 256(x + 4)/x = 0
256x - 256(x + 4) = 0
256x - 256x - 1024 = 0
-1024 = 0
This equation has no real solutions.

Since the equation has no real solutions, it means that there is no minimum area within the given constraints. However, we can still find the possible dimensions by solving the original equation x * y = 256.

Substituting one variable in terms of another:
y = 256 / x

We can now find the dimensions:

Length = x + 4 = x + 4
Width = y + 6 = 256 / x + 6

As there is no minimum area within the given constraints, the overall dimensions that will minimize the amount of paper used do not exist.