A cylindrical metal can with an open top is to be constructed. the can must have a capacity of 24pie cubic inches. the metal used to contruct the bottom of the can costs 3 times as much as the metal in the rest of the can. Find the dimensions(radius and height) of the can that will minimize the cost of its construction.
To find the dimensions of the can that will minimize the cost of its construction, we need to create a cost function and then find the minimum of that function.
Let's first consider the dimensions of the can. The open top indicates that we are dealing with a cylinder, so we have a circular base and a height.
Let's assume the radius of the base is "r" inches and the height of the can is "h" inches.
The volume of the cylinder is given by the formula: V = πr²h
In this case, the volume is given as 24π cubic inches: V = 24π.
Now, let's consider the cost of construction. The metal used for the bottom of the can costs 3 times more than the rest of the can. Let's assume the cost per square inch for the rest of the can is "C" dollars, so the cost for the bottom is 3C dollars per square inch.
The cost of constructing the bottom piece is given by the area of the circle (base) multiplied by the cost per square inch: Cost of bottom = πr² * 3C = 3πr²C.
The cost of constructing the rest of the can (without the bottom) is given by the lateral surface area of the cylinder multiplied by the cost per square inch: Cost of the rest = 2πrh * C = 2πr * h * C.
The total cost of construction is the sum of the cost of the bottom and the cost of the rest: Total cost = Cost of bottom + Cost of the rest.
To minimize the total cost, we need to find the values of r and h that minimize the total cost.
Now, let's express the total cost in terms of a single variable using the volume equation: V = 24π.
From the volume equation, we can solve for h in terms of r: h = 24 / (πr²).
Substituting this expression for h into the cost equation, we get: Total cost = 3πr²C + 2πr(24 / (πr²)) * C = 3πr²C + (48 / r) * C.
Now, we have the total cost expressed in terms of a single variable, r. To find the minimum cost, we need to differentiate the total cost equation with respect to r and set the derivative equal to zero.
d(Total cost)/dr = 6πrC - (48C / r²) = 0.
Simplifying the equation, we get: 6πrC - (48C / r²) = 0.
Solving for r, we find: r = √(8/π).
Substituting this value of r into the volume equation, we can find h: h = 24(π(8/π)) / (π(8/π)²) = 24(π) / (8/π) = 9π.
Therefore, the dimensions of the can that will minimize the cost of its construction are: radius = √(8/π) inches and height = 9π inches.
Please note that we have rounded the values to two decimal places for simplicity.