A rectangular cardboard poser is to have 150 sq ft in for printed matter. It is to have a 3inch margin at the top and bottom and a 2inch margin at the right and left. Find the dimensions of the poser so that the amount of cardboard used is minimized.

if the cardboard has height x and width y,

a = xy

printed area is (x-6)(y-4) = 150

so, we want to minimize xy.

a = xy = x(150/(x-6) + 4)
da/dx = 4(x^2-12x-189)/(x-6)^2

since the denominator is never zero, da/dx=0 when
x^2 - 12x - 189 = 0
(x+9)(x-21) = 0

so, x = 21
y = 14

area of poster is 294
area of printing is 15x10 = 150

To minimize the amount of cardboard used, we need to find the dimensions of the poster that maximize the area of printed matter. Let's denote the width of the printed matter as x inches.

The total width of the cardboard will be x + 2 inches for the left margin and 2 inches for the right margin, so the width of the cardboard is x + 4 inches.

Similarly, the total height of the cardboard will be the height of the printed matter plus 3 inches for the top margin and 3 inches for the bottom margin. The height of the cardboard is x + 6 inches.

To find the area of the printed matter, we multiply the width by the height: Area = (x + 4)(x + 6).

However, we are given that the area of the printed matter is 150 sq ft. Since 1 square foot is equal to 144 square inches, the area of the printed matter in square inches is 150 * 144 = 21600 sq in.

Setting the area of the printed matter equal to 21600 and solving for x:

(x + 4)(x + 6) = 21600

Expanding the equation:

x^2 + 10x + 24 - 21600 = 0

Rearranging the terms:

x^2 + 10x - 21576 = 0

We can solve this quadratic equation using the quadratic formula:

x = (-10 ± √(10^2 - 4 * 1 * (-21576))) / (2 * 1)
x = (-10 ± √(100 + 86304)) / 2
x = (-10 ± √86404) / 2
x = (-10 ± 294.03) / 2

Taking the positive value of x:

x = (-10 + 294.03) / 2
x = 284.03 / 2
x ≈ 142.02

Since the width of the printed matter cannot be negative, we discard the negative value.

Therefore, the width of the printed matter (and the width of the cardboard) is approximately 142.02 inches.

To find the height of the printed matter (and the height of the cardboard), we substitute x back into one of the formulas:

Height = x + 6
Height = 142.02 + 6
Height ≈ 148.02 inches

Therefore, the dimensions of the poster that minimize the amount of cardboard used are approximately 142.02 inches wide and 148.02 inches high.

To find the dimensions of the poster that minimize the amount of cardboard used, we need to consider the area of the cardboard being used.

Let's assume the length of the poster is "L" inches and the width is "W" inches. We can calculate the area of the printed matter by subtracting the margins from the total dimensions.

The length of the printed matter would be L - 2 inches on both sides (2 inches margin on each side), and the width would be W - 3 inches at the top and bottom (3 inches margin on each top and bottom).

So, the area of the printed matter is (L - 2) * (W - 6) square inches.

We are given that the area of the printed matter should be 150 square feet, which is equal to 150 * 144 square inches (since 1 square foot is equal to 144 square inches).

Therefore, we have the equation:

(L - 2) * (W - 6) = 150 * 144

Now, we need to minimize the amount of cardboard used, which is the total area of the poster. The total area of the poster, including the margins, is L * W square inches.

To minimize the amount of cardboard, we need to minimize the total area. So, we need to find the dimensions (L and W) that satisfy the equation (L - 2) * (W - 6) = 150 * 144 and minimize the product L * W.

To solve this problem, we can use calculus or make some observations. Let's try making some observations:

Since we are trying to minimize the total area, we can assume that L > W. This is because, if we have a larger length and a smaller width, we can generally minimize the area.

Now, let's consider the equation (L - 2) * (W - 6) = 150 * 144. Since both L and W are positive values, to get a large product (L-2)*(W-6), we need both L-2 and W-6 to be large. Therefore, we can choose L and W to be as close to the values of 2 and 6 as possible while still satisfying the equation.

By observation, we can assign L = 2 and W = 6. Plugging these values into the equation, we get:

(2 - 2) * (6 - 6) = 0 * 0 = 0

This means that when L = 2 and W = 6, the amount of cardboard used is minimized.

Therefore, the dimensions of the poster that minimize the amount of cardboard used are 2 inches by 6 inches.