you are constructing an open - cylindrical bed for a pet. the material for the bottom of the bed costs $6…

you are constructing an open - cylindrical bed for a pet. the material for the bottom of the bed costs $6 per square foot, and the material for the wall is $3 per square foot. you need the cylindrical bed to have a volume of 10 cubic feet. find the radius and height of the bed that will minimize cost.

you are constructing an open - cylindrical bed for a pet. the material for the bottom of the bed costs $6 per square foot, and the material for the wall is $3 per square foot. you need the cylindrical bed to have a volume of 10 cubic feet. find the radius and height of the bed that will minimize cost.

Answer

Explanation:

Step1: Define variables and formulas

Let $r$ be the radius and $h$ be the height of the cylindrical bed. The volume of the cylinder is $V = \pi r^2 h$. The area of the bottom is $A_{bottom} = \pi r^2$. The area of the wall (lateral surface area) is $A_{wall} = 2 \pi r h$.

Step2: Formulate the cost function

The cost of the bottom material is $6 per square foot, and the wall material is $3 per square foot. Total Cost $C = 6 \cdot A_{bottom} + 3 \cdot A_{wall}$ $C = 6 \pi r^2 + 3 (2 \pi r h)$ $C = 6 \pi r^2 + 6 \pi r h$

Step3: Use the volume constraint

The volume is given as $V = 10$ cubic feet. So, $\pi r^2 h = 10$. Express $h$ in terms of $r$: $h = \frac{10}{\pi r^2}$.

Step4: Substitute h into the cost function

Substitute the expression for $h$ into the cost function to make $C$ a function of $r$ only. $C(r) = 6 \pi r^2 + 6 \pi r \left(\frac{10}{\pi r^2}\right)$ $C(r) = 6 \pi r^2 + \frac{60}{r}$

Step5: Find the derivative of the cost function

To minimize the cost, find the derivative of $C(r)$ with respect to $r$ and set it to zero. $C'(r) = \frac{d}{dr} (6 \pi r^2 + 60 r^{-1})$ $C'(r) = 12 \pi r - 60 r^{-2}$ $C'(r) = 12 \pi r - \frac{60}{r^2}$

Step6: Set the derivative to zero and solve for r

$12 \pi r - \frac{60}{r^2} = 0$ $12 \pi r = \frac{60}{r^2}$ $12 \pi r^3 = 60$ $r^3 = \frac{60}{12 \pi}$ $r^3 = \frac{5}{\pi}$ $r = \left(\frac{5}{\pi}\right)^{1/3}$

Step7: Calculate the corresponding height h

Substitute the value of $r$ back into the equation for $h$: $h = \frac{10}{\pi r^2}$ $h = \frac{10}{\pi \left(\left(\frac{5}{\pi}\right)^{1/3}\right)^2}$ $h = \frac{10}{\pi \left(\frac{5}{\pi}\right)^{2/3}}$ $h = \frac{10}{\pi^{1/3} 5^{2/3}}$ $h = \frac{2 \cdot 5}{\pi^{1/3} 5^{2/3}}$ $h = \frac{2 \cdot 5^{1/3}}{\pi^{1/3}}$ $h = 2 \left(\frac{5}{\pi}\right)^{1/3}$

Answer:

The radius that minimizes cost is $r = \left(\frac{5}{\pi}\right)^{1/3}$ feet. The height that minimizes cost is $h = 2 \left(\frac{5}{\pi}\right)^{1/3}$ feet.