1) A rectangular page is to contain 24 square inches of print. The page has to have a 6-inch margin on top and at the bottom and a 1-inch margin on each side. Find the dimensions of the page that minimize the amount of paper used.
2) A cable runs along a wall from C to P at a cost of $7 per meter, and straight from P to M at a cost of $25 per meter. Let x be the distance from C to P. If M is 48 meters from the nearest point A on the wall where P lies, and A is 106 meters from C, find x such that the cost of installing the cable is minimized and find this cost.
If the paper is x by y inches, then the printed area is
(x-2)(y-12)
The paper has area
a = xy = x(24/(x-2) + 12) = 12x^2/(x-2)
da/dt = 12x(x-4)/(x-2)^2
da/dt=0 when x=4
So, the printed area is 4x6 and the page is 6x18
Now see what you can do with the other problem
To find the dimensions of the page that minimize the amount of paper used, we need to maximize the area of the printed material while keeping the margins constant.
Let's assume the dimensions of the printed material are width and height, denoted as x and y, respectively. The total area of the printed material can be calculated as (x - 2) * (y - 12) since each side has a 1-inch margin, which reduces the effective width by 2 inches, and the top and bottom margins are 6 inches each.
Since we want to minimize the amount of paper used, we need to maximize the area of the printed material. Therefore, we can rewrite the area formula as:
A = (x - 2) * (y - 12)
To find the dimensions that minimize the amount of paper used, we can find the critical points of this function. The critical points occur when the partial derivatives with respect to x and y are both equal to zero.
Now, let's differentiate the area formula with respect to x:
dA/dx = y - 12
Setting dA/dx = 0, we get:
y - 12 = 0
y = 12
Next, let's differentiate the area formula with respect to y:
dA/dy = x - 2
Setting dA/dy = 0, we get:
x - 2 = 0
x = 2
Therefore, the dimensions of the page that minimize the amount of paper used are 2 inches by 12 inches.
For the second problem, to find the distance x that minimizes the cost of installing the cable, we need to analyze the cost function.
Let's assume that the cost of installing the cable from C to P is C_CP and the cost of installing the cable from P to M is C_PM.
The cost of installing the cable from C to P is $7 per meter, so C_CP = 7x.
The cost of installing the cable from P to M is $25 per meter, so C_PM = 25(48 - x).
The total cost function, C(x), can be written as:
C(x) = C_CP + C_PM
C(x) = 7x + 25(48 - x)
To find the distance x that minimizes the cost, we need to analyze the critical points of the cost function. The critical point occurs when the derivative of the cost function with respect to x is equal to zero.
Now, let's differentiate the cost function with respect to x:
dC(x)/dx = 7 - 25
Setting dC(x)/dx = 0, we get:
7 - 25 = 0
-18 = 0
Since -18 is not equal to zero, we don't have a critical point.
However, we have a boundary point constraint where x cannot be greater than 48 (the distance from A to M). Therefore, we need to evaluate the cost function at the endpoints.
When x = 0 (distance from C to P), the cost is C_CP = 7(0) = $0.
When x = 48 (distance from P to M), the cost is C_PM = 25(48 - 48) = $0.
Since the cost is the same at both endpoints, the distance x doesn't matter in terms of cost. Both endpoints result in zero cost.
Therefore, the cost of installing the cable is minimized when the distance from C to P is 0 meters, and the total cost is $0.
1) To minimize the amount of paper used, we need to minimize the area of the page.
Let's assume the length of the page is L and the width is W.
We know that there is a 6-inch margin on the top and bottom, so the printable area in height is L - 2(6) = L - 12 inches.
Similarly, there is a 1-inch margin on each side, so the printable area in width is W - 2(1) = W - 2 inches.
Since the area of a rectangle is given by A = length x width, we can express the area of the page as:
A = (L - 12) x (W - 2)
We are given that the area of print is 24 square inches, so we can set up the equation:
24 = (L - 12) x (W - 2)
To minimize the amount of paper used, we need to minimize the area A. One way to do this is to find the critical points of A by taking the partial derivatives with respect to L and W and setting them equal to zero.
Let's differentiate the equation with respect to L:
dA/dL = W - 2 = 0
Solving this equation, we find W = 2.
Now, let's differentiate the equation with respect to W:
dA/dW = L - 12 = 0
Solving this equation, we find L = 12.
Therefore, the dimensions of the page that minimize the amount of paper used are L = 12 inches and W = 2 inches.
2) To minimize the cost of installing the cable, we need to find the distance from C to P, denoted as x.
Given that M is 48 meters from the nearest point A on the wall where P lies, and A is 106 meters from C, we can create the following diagram:
C - A - P - M
Since we are given the distances, we can write the following equations:
AC = 106 meters
AM = 48 meters
By using the Pythagorean theorem, we can find the distance CM:
CM^2 = AC^2 - AM^2
CM^2 = 106^2 - 48^2
CM^2 = 11236 - 2304
CM^2 = 8932
CM = √8932
CM ≈ 94.56 meters
Now, let's find the cost of installing the cable. The cost from C to P is $7 per meter, and the cost from P to M is $25 per meter.
The total cost is given by the equation:
cost = (CP * $7) + (PM * $25)
cost = (x * $7) + ((CM - x) * $25)
We want to minimize the cost, so we need to find the value of x that minimizes this equation.
To find the critical point, we differentiate the cost equation with respect to x and set it equal to zero:
d(cost)/dx = 7 - 25 = 0
Solving this equation, we find x = 7/25.
Therefore, to minimize the cost of installing the cable, the distance from C to P should be approximately 7/25 meters, and the cost would be:
cost = ((7/25) * $7) + ((94.56 - 7/25) * $25)