The top and bottom margins of a poster are 8 cm and the side margins are each 9 cm. If the area of printed material on the poster is fixed at 382 square centimeters, find the dimensions of the poster with the smallest area.
If the printed area has width x and height y,
xy=382
The area of the poster is
a = (x+18)(y+16)
= (x+18)(382/x + 16)
= 16x + 670 + 6876/x
For minimum area, da/dx=0, so
16 - 6876/x^2 = 0
x = 3/2 √191
Now you can figure the poster dimensions.
Let's assume that the width of the printed material is x cm.
The total width of the poster will be the width of the printed material plus the left and right margins, which is (x + 9 + 9) = (x + 18) cm.
Similarly, the total height of the poster will be the height of the printed material plus the top and bottom margins, which is (382/x + 8 + 8) = (382/x + 16) cm.
To find the dimensions of the poster with the smallest area, we need to minimize the area of the poster. The area of the poster can be calculated by multiplying its width and height:
Area = (x + 18) * (382/x + 16)
To find the minimum area, we can take the derivative of the area equation with respect to x and set it equal to zero:
d(Area)/dx = 0
Differentiating the area equation, we get:
(x + 18)' * (382/x + 16) + (x + 18) * (382/x + 16)' = 0
Simplifying and solving for x:
1 * (382/x + 16) + (x + 18) * (-382/x^2) = 0
382/x + 16 - 382(x + 18)/x^2 = 0
382/x + 16 - 382x/x^2 - 382*18/x^2 = 0
382 + 16x - 382x - 382*18 = 0
16x - 382x = 382*18 - 382
-366x = 382*18 - 382
x = (382*18 - 382) / -366
x ≈ 3.89
Now that we have the value of x, we can calculate the width and height of the poster:
Width = x + 18 ≈ 3.89 + 18 ≈ 21.89 cm
Height = 382/x + 16 ≈ 382/3.89 + 16 ≈ 99.51 cm
Therefore, the dimensions of the poster with the smallest area are approximately 21.89 cm by 99.51 cm.
To find the dimensions of the poster with the smallest area, we can set up an equation and optimize it. Let's assume the length of the poster is L and the width of the poster is W.
From the given information, we know that the top and bottom margins are both 8 cm. So the effective length of the printed material will be L-2*8 = L-16.
Similarly, the side margins are each 9 cm. So the effective width of the printed material will be W-2*9 = W-18.
The area of the printed material is fixed at 382 square centimeters. This can be given as:
(L-16) * (W-18) = 382
Now, we need to minimize the area of the poster, which is given by A = L * W.
We can rewrite the area expression in terms of L, W:
A = (L-16) * (W-18) + 2 * 8 * (W-18) + 2 * 9 * (L-16)
Simplifying the equation:
A = L * W - 16W - 18L + 288 + 16W - 288 + 18L - 288
A = L * W - 16W - 18L + 16W - 18L
A = L * W - 34L - 34W
Now, we need to find the values of L and W that minimize the area A. To do this, we can take partial derivatives of A with respect to L and W, and set them equal to zero:
∂A/∂L = W - 34 = 0 (partial derivative with respect to L)
∂A/∂W = L - 34 = 0 (partial derivative with respect to W)
Solving these equations, we get L = W = 34.
Therefore, the dimensions of the poster with the smallest area are 34 cm for both the length and width.