The top and bottom margins of a poster are 6 cm and the side margins are each 7 cm. If the area of printed material on the poster is fixed at 384 square centimeters, find the dimensions of the poster with the smallest area.

let length of printed area be x

let the width be y

so xy = 384
y = 384/x

length of paper = x+12
width of paper = y+14

area of paper = (x+12)(y+14)
= xy + 14x + 12y + 168
= x(384/x) + 14x + 14(384/x) + 168
= 384 + 14x + 5376/x + 168

d(area)/dx = 14 - 5376/x^2
= 0 for min of area

5376/x^2 = 14
14x^2 = 5376
x^2 = 384
x = √384 = appr 19.596 cm
then y = 19.596

poster is 19.596+12 by 19.596+14
etc

Wrong answer

To find the dimensions of the poster with the smallest area, we can start by finding the dimensions of the printed material.

Let's assume that the width of the printed material is x cm. Since there are side margins of 7 cm on each side, the total width of the poster is (x + 2(7)) cm.

Similarly, let's assume that the height of the printed material is y cm. Since there are top and bottom margins of 6 cm each, the total height of the poster is (y + 2(6)) cm.

The area of the printed material is given as 384 square cm. So, we have the equation:

(x)(y) = 384

To minimize the area of the poster, we need to minimize the product xy. Let's rewrite the equation in terms of one variable:

y = 384/x

Now, let's substitute this value of y in the expression for the total height:

(y + 2(6)) = (384/x) + 12

Next, let's substitute this expression for y and the expression for the total width into the formula for the area of the poster:

(x + 2(7))(384/x + 12) = Area

To find the dimensions of the poster with the smallest area, we need to find the critical points of the area function. To do this, we can differentiate the area function with respect to x, set it to zero, and solve for x.

Differentiating the area function, we have:

d(Area)/dx = (384/x + 12)(2) - (x + 14)(384/x^2)

Setting this to zero:

(384/x + 12)(2) - (x + 14)(384/x^2) = 0

Simplifying and rearranging the equation:

(2 * 384 + 24x) - (x + 14)(384/x^2) = 0

(768 + 24x) - (384/x) - 5376/x^2 = 0

Multiplying through by x^2:

768x^2 + 24x^3 - 384x - 5376 = 0

Simplifying further:

8x^3 - 128x^2 + 16x - 224 = 0

We can now solve this cubic equation to find the value of x. However, finding the exact value of x may be difficult, so we can use numerical methods or approximation techniques to find an approximate value of x.

Once we have the value of x, we can substitute it back into the equation for y = 384/x to find the value of y. These values of x and y will give us the dimensions of the poster with the smallest area.

To find the dimensions of the poster with the smallest area, we need to use a mathematical technique called calculus. Specifically, we can use the concept of optimization to solve this problem.

First, let's assume that the width of the printed material is x centimeters. Since the top and bottom margins are both 6 cm, the total height of the poster including margins will be (x + 12) cm. Similarly, the total width of the poster including margins will be (x + 14) cm, as the side margins are each 7 cm.

The area of the poster can be expressed as the product of its height and width: A = (x + 12)(x + 14).

Given that the area of the printed material is 384 square centimeters, we can set up the following equation:

A = (x + 12)(x + 14) = 384.

To find the dimensions of the poster with the smallest area, we need to find the value of x that minimizes the area A.

To do this, we can start by expanding the equation:

A = x^2 + 26x + 168.

Next, we can find the derivative of A with respect to x:

A' = 2x + 26.

To find the minimum, we set the derivative equal to zero and solve for x:

2x + 26 = 0,
2x = -26,
x = -13.

But since the width of the printed material cannot be negative, we discard this solution.

Therefore, there is no minimum for the area of the poster. In this case, the smallest area is achieved when the width of the printed material approaches zero. This means that the dimensions of the poster with the smallest area would have no printed material, resulting in just the margins.

Therefore, if the area of the printed material is fixed at 384 square centimeters, the dimensions of the poster with the smallest area would be 6 cm for the top and bottom margins, and 7 cm for each side margin.