A small dog shelter in the shape of a rectangular solid is to have a square base and an open front. If the volume is to be 36ft^3, find the dimensions for which the amount of material needed is a minimum. V=2x*y
Your equation is wrong.
If x is the base dimension and y is the height,
V = x^2*y = 36 , so
y = 36/x^2
The amount (area) of material needed is:
A = 2x^2 + 3 x*y
= 2x^2 + 108/x
dA/dx = 0 @ minimum area
4x = 108/x^2
x^3 = 27
x = 3 feet
y = 36/(3^2) = 4 feet
I hate calculus
Try to see the beauty of it.
Without calculus, you'd get hit by a car whenever you cross a street. When you step off the curb, you have seen that the approaching car is growing in your field of vision at a rate that indicates it will not arrive at the crosswalk by the time you get to the other side, moving at your known walking speed.
You don't consciously assign values to these variables, but your brain takes care of it for you. When you were small, your parents snatched you back until you learned how to interpret what you saw.
Calculus is such a useful tool, you ought to look forward to mastering the techniques it provides to solve real-life problems.
To find the dimensions for which the amount of material needed is at a minimum, we can use the derivative of the volume function.
Given:
Volume of the shelter (V) = 36 ft^3
Volume formula: V = 2xy, where x and y are the dimensions of the rectangular solid.
1. Write the volume equation in terms of one variable.
Since we want to find the dimensions that minimize the amount of material (which is equivalent to minimizing the surface area), we know that two sides of the base and all four sides of the rectangular solid contribute to the material needed.
For a square base, two sides have dimensions x and the other two sides have dimensions y. The third side, which is the height of the rectangular solid, is not part of the surface area.
The equation for the amount of material needed is then A = x*x + y*x + y*x + x*y (summing up the four sides).
2. Simplify the equation:
A = 2xy + 2xy + x^2
A = 4xy + x^2
3. Substitute the volume equation into the material equation:
A = 4xy + (V/2y)^2
4. Differentiate the material equation:
To find the minimum of the material equation, we need to differentiate it with respect to x and set it equal to zero.
dA/dx = 4y - (V/y^2) = 0
5. Solve for y:
4y - (V/y^2) = 0
4y = (V/y^2)
4y^3 = V
y^3 = V/4
y = (V/4)^(1/3)
6. Substitute the value of y back into the volume equation to find the corresponding x value:
V = 2x * [(V/4)^(1/3)]
7. Simplify and solve for x:
V/2 = x * [(V/4)^(1/3)]
x = (V/2) / [(V/4)^(1/3)]
x = (V/2) * [(4/V)^(1/3)]
x = V^(1/3)
Therefore, the dimensions for which the amount of material needed is a minimum are:
x = V^(1/3)
y = (V/4)^(1/3)