A solid is formed by adjoining two hemispheres to the ends of a right circular cylinder. An industrial tank of this shape must have a volume of 4640 cubic feet. The hemispherical ends cost twice as much per square foot of surface area as the sides. Find the dimensions that will minimize the cost. (Round your answers to three decimal places.)

Answers

Answer 1

Answer:

The dimensions that minimize the cost of building the tank are $r\approx 6.5ft \text{ and } h \approx 26.0ft$ .

Let

$h=\text{the height of the cylindrical sides}\nr=\text{the radius of the hemispheres and the cylindrical sides}$

and

$T=\text{the total cost}\nu_c=\text{the unit cost of building the cylindrical sides}\nu_h=\text{the unit cost of building the hemispherical surfaces}$

and

$S_c=\text{the surface area of the cylinder}\n=2\pi rh\n\nS_h=\text{the total surface area of both hemispheres}\n=4\pi r^2$

Therefore,

$T=u_cSc+u_hSh$

From the question

$u_h=2u_c$

The total cost becomes

$T=u_cSc+2u_cSh\n=2u_c \pi rh+8u_c \pi r^2$

We need to eliminate $h$ . The volume from the question gives a way out

$4640=(4)/(3)\pi r^3 +\pi r^2h\n\nh=(4640)/(\pi r^2)-(4r)/(3)$

substitute into the formula for total cost gives, after simplifying

$T=(16u_c\pi r^2)/(3)+(9280u_c)/(r^2)$

differentiating with respect to $r$ , we get

$(dT)/(dr)=(32)/(3)u_c\pi r-(9280u_c)/(r^2)$

at extrema

$(dT)/(dr)=0\n\n\implies (32)/(3)u_c\pi r=(9280u_c)/(r^2)\n\nr=\sqrt[3]{(870)/(\pi)}\approx 6.5ft$

To confirm that $r$ is a minimum value, carry out the second derivative test

$(d^2T)/(dr^2)=(32u_c\pi)/(3)+(18560)/(r^3)$

substituting $r=\sqrt[3]{(870)/(\pi)}$ , we get that $(d^2T)/(dr^2)$ > $0$ , confirming that minimum value

To find $h$ , recall that

$h=(4640)/(\pi r^2)-(4r)/(3)$

substituting $r$ , we get $h\approx 26.0ft$ as the corresponding minimum height

Therefore, $r\approx 6.5ft \text{ and } h \approx 26.0ft$ minimize the total cost of building the tank.

Learn more about minimizing dimensions to reduce costs here: brainly.com/question/19053049

Answer 2

Answer:

Final answer:

The problem involves finding the dimensions of a cylinder and two hemispheres that minimize the cost to build an industrial tank of a specific volume. This involves setting up equations for the volume and cost, and then using calculus to find the dimensions that minimize the cost.

Explanation:

This problem can be solved using calculus. Let's denote the radius of both the hemispheres and the cylinder as r and the height of the cylinder as h. The total volume of the solid is the sum of the volume of the cylinder and the two hemispheres. Using the formulas for the volumes of a cylinder and hemisphere, we have:

V = (πr²h) + 2*(2/3πr³) = 4640 cubic feet.

The total cost of the material is proportional to the surface area. The surface area of the two hemispheres is twice as expensive as that of the sides of the cylinder, so we have:

Cost = 2*(2πr²) + πrh.

To minimize the cost, we can take the derivative of the Cost function with respect to r and h, set them equal to zero, and solve for r and h.

This problem involves calculus, the volume of cylinders and spheres, and optimization, which are topics covered in high school mathematics.

Learn more about Mathematics Optimization Problems here:

brainly.com/question/32199704

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