Question
An open tank of square base is to be made from a square sheet of metal if sides 20m by cutting square pieces from each of the corners, lifting the resulting flaps and soldering the edges together. Find the maximum capacity of such tank
Answers
To find the maximum capacity of the tank, we need to determine the dimensions of the square base after the corners are cut.
Let x be the length of one side of the square cut from each corner.
The length of the base of the tank will be (20 - 2x) meters, and the height of the tank will also be (20 - 2x) meters.
The volume of a rectangular prism is given by the formula: V = length * width * height.
In this case, the volume of the tank is: V = (20 - 2x) * (20 - 2x) * x.
To find the maximum capacity of the tank, we need to maximize this volume.
Taking the derivative of the volume function with respect to x:
dV/dx = -8x^2 + 80x - 400.
Setting this derivative equal to zero and solving for x:
-8x^2 + 80x - 400 = 0.
Dividing by -8:
x^2 - 10x + 50 = 0.
Using the quadratic formula to solve for x:
x = (-(-10) ± sqrt((-10)^2 - 4(1)(50))) / (2(1)).
x = (10 ± sqrt(100 - 200)) / 2.
x = (10 ± sqrt(-100)) / 2.
Since the square root of a negative number is not a real number, there are no real solutions to this equation.
This means that the volume of the tank is always decreasing as x increases, which implies that the maximum capacity of the tank occurs when x = 0.
When x = 0, the tank is just a square box without any corners cut, so the dimensions of the tank are 20m x 20m x 20m.
Therefore, the maximum capacity of the tank is V = 20m * 20m * 20m = 8,000 cubic meters.
Let x be the length of one side of the square cut from each corner.
The length of the base of the tank will be (20 - 2x) meters, and the height of the tank will also be (20 - 2x) meters.
The volume of a rectangular prism is given by the formula: V = length * width * height.
In this case, the volume of the tank is: V = (20 - 2x) * (20 - 2x) * x.
To find the maximum capacity of the tank, we need to maximize this volume.
Taking the derivative of the volume function with respect to x:
dV/dx = -8x^2 + 80x - 400.
Setting this derivative equal to zero and solving for x:
-8x^2 + 80x - 400 = 0.
Dividing by -8:
x^2 - 10x + 50 = 0.
Using the quadratic formula to solve for x:
x = (-(-10) ± sqrt((-10)^2 - 4(1)(50))) / (2(1)).
x = (10 ± sqrt(100 - 200)) / 2.
x = (10 ± sqrt(-100)) / 2.
Since the square root of a negative number is not a real number, there are no real solutions to this equation.
This means that the volume of the tank is always decreasing as x increases, which implies that the maximum capacity of the tank occurs when x = 0.
When x = 0, the tank is just a square box without any corners cut, so the dimensions of the tank are 20m x 20m x 20m.
Therefore, the maximum capacity of the tank is V = 20m * 20m * 20m = 8,000 cubic meters.
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