101k views
5 votes
An industrial tank of the shape described in Exercise 33 must have a volume of 3000 cubic feet. If the con- struction cost of the hemispherical ends is twice as much per square foot of surface area as the sides, find the dimensions that will minimize cost.

User GaspardP
by
7.8k points

1 Answer

0 votes

Final answer:

To minimize the cost of the industrial tank, we need to determine the dimensions that minimize the surface area of the tank. This can be done by setting up equations for the surface area of the tank and the construction cost and then finding the derivative and solving for the critical points. Considering that the volume of the tank is given, we can solve for one variable in terms of the other and substitute it into the cost function. By finding the derivative of the cost function with respect to one variable, setting it equal to zero, and solving for that variable, we can determine the dimensions that minimize the cost of the tank.

Step-by-step explanation:

In order to minimize the cost, we need to determine the dimensions that will minimize the surface area of the tank. Let's denote the radius of the hemispherical ends as r and the height of the cylindrical part as h. We can then determine the formulas for the surface area of the hemispheres and the surface area of the cylindrical part of the tank.

The surface area of the hemispheres is given by 2πr^2 since we have two hemispherical ends. The surface area of the cylindrical part is given by 2πrh. The volume of the tank is given by the sum of the volumes of the two hemispheres and the volume of the cylindrical part, which is equal to 2/3πr^3 + πr^2h.

We are given that the construction cost of the hemispherical ends is twice as much per square foot of surface area as the sides. Let's denote the cost per square foot of the hemispheres as C and the cost per square foot of the sides as 2C. The total cost of the tank is then given by the cost of the hemispheres plus the cost of the sides, which is equal to 2πr^2C + 2πrh(2C).

Our goal is to minimize this cost function. We can do this by taking its derivative with respect to r and h, setting them equal to zero, and solving for r and h. However, since the problem specifies that the volume of the tank is 3000 cubic feet, we can use this information to solve for h in terms of r and substitute it into the cost function.

User AAlvz
by
7.1k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.