Question

You have been asked to design a can shaped like right circular cylinder that can hold a volume of 432π-cm3. What dimensions of the can (radius and height) will use the least amount of material?(Hint: The lightest can will be the can with the smallest surface area.)

Answer 1

Answer:

$Height = 12cm$

$Radius = 6cm$

Step-by-step explanation:

Given

Represent volume with v, height with h and radius with r

$V = 432\pi$

Required

Determine the values of h and r that uses the least amount of material

Volume is calculated as:

$V = \pi r^2h$

Substitute 432π for V

$432\pi = \pi r^2h$

Divide through by π

$432 = r^2h$

Make h the subject:

$h = \frac{432}{r^2}$

Surface Area (A) of a cylinder is calculated as thus:

$A=2\pi rh+2\pi r^2$

Substitute $\frac{432}{r^2}$ for h in $A=2\pi rh+2\pi r^2$

$A=2\pi r(\frac{432}{r^2})+2\pi r^2$

$A=2\pi (\frac{432}{r})+2\pi r^2$

Factorize:

$A=2\pi (\frac{432}{r} + r^2)$

To minimize, we have to differentiate both sides and set $A' = 0$

$A'=2\pi (-\frac{432}{r^2} + 2r)$

Set $A' = 0$

$0=2\pi (-\frac{432}{r^2} + 2r)$

Divide through by $2\pi$

$0= -\frac{432}{r^2} + 2r$

$\frac{432}{r^2} = 2r$

Cross Multiply

$2r * r^2 = 432$

$2r^3 = 432$

Divide through by 2

$r^3 = 216$

Take cube roots of both sides

$r = \sqrt[3]{216}$

$r = 6$

Recall that:

$h = \frac{432}{r^2}$

$h = \frac{432}{6^2}$

$h = \frac{432}{36}$

$h = 12$

Hence, the dimension that requires the least amount of material is when

$Height = 12cm$

$Radius = 6cm$