1) A rectangular page is to contain 16 square inches of print. The page has to have a 2-inch margin on top and at the bottom and a 2-inch margin on each side. Find the dimensions of the page that minimize the amount of paper used.
2) A rectangular garden of area 480 square feet is be surrounded on three sides by a brick wall costing $12 per foot and on one side by a fence costing $8 per foot. Find the dimensions of the garden such that the cost of the materials is minimized.
As with most such problems, I expect you will find that the maximum area occurs on a square page. But, let's check it out.
If the dimensions are x and y, we have
(x-4)(y-4) = 16
a = xy = x*(16/(x-4) + 4)
da/dx = 4x(x-8)/(x-4)^2
da/dx = 0 when x=8
So, the page is 8x8, with the printed area 4x4, having an area of 16
It would have been a better problem had the margins not been the same. Just sayin'
If the fenced side is y and the wall sides are x, then we have
xy = 480
and the cost is
c = 12*2x + 8y = 24x + 8(480/x) = 24x + 3840/x
dc/dx = 24(x^2-160)/x^2
dc/dx = 0 when x = 4√10
So, the garden is 4√10 by 120/√10
1) To find the dimensions of the page that minimize the amount of paper used, we need to minimize the total area of the page.
Let's assume the width of the page is x inches. The length of the page will be (16 / x) inches because the area of a rectangle is given by length times width (16 square inches = length x width).
Considering the 2-inch margins on both sides of the page, the effective width will be (x - 4) inches. Similarly, considering the top and bottom margins, the effective length will be (16 / (x - 4) - 4) inches.
The total area of the page can be calculated as the product of the effective length and effective width:
Total area = (x - 4) * (16 / (x - 4) - 4)
To find the dimensions that minimize the total area, we can take the derivative of the total area with respect to x, set it equal to zero, and solve for x.
Differentiating the total area function:
d(Total area) / dx = ((16 / (x - 4) - 4) - 4) - ((x - 4) * (-16 / (x - 4)^2))
Setting the derivative equal to zero:
0 = ((16 / (x - 4) - 4) - 4) - ((x - 4) * (-16 / (x - 4)^2))
Simplifying the equation:
0 = (16 / (x - 4) - 4) - 4 + (16 / (x - 4))
0 = 16 / (x - 4)
Solving for x:
16 = x - 4
x = 20
Therefore, the width of the page that minimizes the amount of paper used is 20 inches.
To find the length of the page, we can substitute the value of x back into the equation for effective length:
Effective length = 16 / (x - 4) - 4
Effective length = 16 / (20 - 4) - 4
Effective length = 2 inches
Therefore, the dimensions of the page that minimize the amount of paper used are 20 inches by 2 inches.
2) To find the dimensions of the garden that minimize the cost of materials, we need to minimize the total cost of the materials.
Let's assume the length of the garden is L feet and the width is W feet. The area of the garden is given by length times width (L x W = 480 square feet).
Considering that three sides of the garden are surrounded by a brick wall and one side by a fence, the total cost of the materials can be calculated as follows:
Total cost = (3 * perimeter of the garden with brick wall * $12 per foot) + (perimeter of the garden with fence * $8 per foot)
The perimeter of the garden with brick wall can be calculated as follows:
Perimeter of garden with brick wall = 2L + W
The perimeter of the garden with fence can be calculated as follows:
Perimeter of garden with fence = L
Substituting these values into the equation for the total cost:
Total cost = (3 * (2L + W) * $12) + (L * $8)
Total cost = (72L + 36W) + 8L
Total cost = 80L + 36W
To minimize the total cost, we need to minimize the total cost function with respect to L and W. Taking the partial derivatives of the total cost function:
∂Total cost / ∂L = 80
∂Total cost / ∂W = 36
Since the partial derivatives are constants, they do not depend on L or W. Therefore, the dimensions of the garden do not affect the total cost.
There is no specific set of dimensions that will minimize the cost of materials. Any garden with an area of 480 square feet will have the same total cost, regardless of its dimensions.
To solve these optimization problems, we need to use calculus. Let's start with the first problem:
1) The goal is to find the dimensions of the rectangular page that minimize the amount of paper used.
Let's assume the length of the page is L and the width is W. We can set up an equation for the area of the page:
Area of the page = L * W
We know that the page has a 2-inch margin on top, bottom, left, and right. This means the printable area is (L - 4) inches by (W - 4) inches.
Area of the printable area = (L - 4) * (W - 4)
We are given that the page needs to contain 16 square inches of print. So we can set up another equation:
(L - 4) * (W - 4) = 16
Now, we have two equations:
Equation 1: Area of the page = L * W
Equation 2: (L - 4) * (W - 4) = 16
We need to find the dimensions L and W that minimize the amount of paper used. To do this, we can find the minimum value of the area by taking the derivative of the area equation with respect to L and W, and then setting those derivatives equal to zero.
Let's first solve Equation 2 for L:
L = 4 + 16/(W - 4)
Now we can substitute this into Equation 1:
Area of the page = (4 + 16/(W - 4)) * W = 4W + 16 - 64/(W - 4)
To find the minimum, we differentiate the area equation with respect to W:
d(Area)/dW = 4 + 64/(W - 4)^2
To find the critical points, set d(Area)/dW = 0:
4 + 64/(W - 4)^2 = 0
Solve for W: 64/(W - 4)^2 = -4
(W - 4)^2 = -64/4 = -16
Since the square of a real number cannot be negative, there are no real solutions for W. This means the minimum point occurs either at an extreme value of W or at the endpoints of the interval.
To investigate further, we can calculate the area when W = 8, 10, and 12 (the endpoints and a value in between):
When W = 8, L = 20
Area of the page = 20 * 8 = 160 square inches
When W = 10, L = 12
Area of the page = 12 * 10 = 120 square inches
When W = 12, L = 8
Area of the page = 8 * 12 = 96 square inches
Based on these calculations, we can conclude that the dimensions of the page that minimize the amount of paper used are W = 12 inches and L = 8 inches.