The top and bottom margins of a poster are 2 cm and the side margins are each 8 cm. If the area of printed material on the poster is fixed at 380 square centimeters, find the dimensions of the poster with the smallest area.
Width =
Height =
poster dimensions: L+4; W+16
LW=380
areatotal=(L+4)(W+16)=(L+4)(380/L+ 16)
dareatotal/dl=0=(380/L + 16) + (L+4)(-380/L^2)
solve for L
0= 380L+16*L^2- 380L-4*380
L^2=380/4=95
L= sqrt 95
at this point recheck my math work, I have a headache. If correct, solve for W, then solve for the poster dimensions.
wh = 380
a = (w+16)(h+4)
= (w+16)(380/w + 4)
= 4w + 444 + 6080/w
da/dw = 4 - 6080/w^2
minimum area when da/dw = 0. So solve that and evaluate w and h. Then the poster dimensions are w+16 and h+4.
To find the dimensions of the poster with the smallest area, we can define the width and height of the printed material on the poster. Let's denote the width as "x" and the height as "y".
Given that the top and bottom margins are 2 cm each, and the side margins are 8 cm each, we can express the dimensions of the poster as:
Width of the poster: x + 2(8) = x + 16 cm
Height of the poster: y + 2(2) = y + 4 cm
The area of the printed material on the poster is fixed at 380 square centimeters, so we can set up the following equation:
(x + 16)(y + 4) = 380
Now, let's solve this equation to find the dimensions of the poster with the smallest area.
First, we can simplify the equation:
xy + 4x + 16y + 64 = 380
Rearranging the equation:
xy + 4x + 16y = 316
Now, let's solve for y in terms of x:
y = (316 - 4x) / (x + 16)
To find the dimensions of the poster with the smallest area, we need to minimize the area function in terms of x. The area function is given by:
Area = x * y
Substituting the expression for y in terms of x, we get:
Area = x * (316 - 4x) / (x + 16)
To find the x-value that minimizes the area, we can take the derivative of the area function with respect to x, set it equal to zero, and solve for x:
d(Area) / dx = (316 - 4x) / (x + 16) - [(x * (-4)) / (x + 16)^2] = 0
Simplifying the equation and solving for x:
(316 - 4x) / (x + 16) + (4x / (x + 16)^2) = 0
316 - 4x + 4x = 0
316 = 0
Therefore, there is no x-value that satisfies the equation, which means that the area cannot be minimized. This indicates that there is no smallest area possible for the poster under the given constraints.
To find the dimensions of the poster with the smallest area, let's set up an equation using the given information.
Let the width of the printed material be x and the height of the printed material be y.
According to the given information, the top and bottom margins are 2 cm each, so the total height of the poster (including the margins) is y + 2 + 2 = y + 4 cm.
Similarly, the side margins are 8 cm each, so the total width of the poster is x + 8 + 8 = x + 16 cm.
Using the formula for the area of a rectangle (A = length * width), we can set up the following equation:
Area = (x + 16) * (y + 4) = 380
Now, let's solve this equation to find the dimensions of the poster with the smallest area.
Divide both sides of the equation by (y + 4):
(x + 16) * (y + 4) / (y + 4) = 380 / (y + 4)
Simplify:
x + 16 = 380 / (y + 4)
Now, let's isolate y by subtracting 16 from both sides:
x = 380 / (y + 4) - 16
To minimize the area, we need to find the minimum value of x as y varies.
Now, we can find the derivative of x with respect to y and set it to zero:
d(x) / d(y) = -380 / (y + 4)^2 = 0
Solve for y:
-380 / (y + 4)^2 = 0
Since a fraction is equal to zero only when the numerator is zero, we have:
-380 = 0
This is not possible, so there is no minimum value for x.
Therefore, there is no dimensions of the poster that will result in the smallest area of 380 square centimeters.