A retangular poster is to be made out of cardboard. The margins on the top and the bottom will be 2 inches. The margins on the sides will be 1 inch. The area of the poster devoted to the printed material must be 300 square inches.

(a) to the nearest .001 inches, find the dimensions of the poster that use the least amount of cardboard.

(b) To the nearest .001 square inches, find the area of the poster that uses the least amount of cardboard.

I know the Area of a rectange = Length X Width. But for part a basically it's converting square inches to inches

300

To find the dimensions of the poster that use the least amount of cardboard, we need to determine the length and width of the printed area. Let's solve part (a) of the problem first.

Step 1: Determine the dimensions of the printed area.
The total area of the poster is the sum of the printed area and the margins.
Let's denote the length of the printed area as L and the width as W.

Considering the margins, the length of the poster is:
Length = L + 2 inches (top margin) + 2 inches (bottom margin)

The width of the poster is:
Width = W + 1 inch (left margin) + 1 inch (right margin)

Step 2: Express the total area in terms of L and W.
The total area of the cardboard used for the poster is given by:
Total area = Length × Width

Total area = (L + 4) × (W + 2)

Step 3: Express the printed area in terms of L and W.
The area of the printed material is given as 300 square inches.

Printed area = L × W

Step 4: Substitute the expressions from step 2 and step 3 into the total area equation.
Total area = (L + 4) × (W + 2)

300 = L × W

Step 5: Solve for L or W in terms of the other variable.
From the area equation, we can solve for one variable and substitute it back into the total area equation to find the other variable.

Let's solve for W in terms of L from the printed area equation:
W = (300 / L)

Now substitute this expression for W in the total area equation:
Total area = (L + 4) × ((300 / L) + 2)

Step 6: Find the minimum total area by taking the derivative.
To find the dimensions that use the least amount of cardboard, we need to minimize the total area equation.

Differentiate the total area equation with respect to L:

d(Total area) / dL = 0

Now solve for L:

(L + 4) × (300 / L + 2) = 0

Solving this equation may require more advanced techniques or numerical methods. Once you find the value(s) of L, substitute it back into the equation W = (300 / L) to get the corresponding width.

For part (b), once you have found the dimensions of the poster that use the least amount of cardboard from part (a), use the formula for the area of a rectangle (Area = Length × Width) with these dimensions to find the area to the nearest .001 square inches.