Answer to Question #159472 in Calculus for Sai preeth varma Lolakpuri

Question

A rectangular box at the top to have volume of 32cubic feet. Find the dimensions of the box requiring least material for it"s construction

Answer 1

Step-by-step explanation:

Volume = LBH = 32 cubic feet

To have minimum surface area L = B

$\rightarrow$ Volume = L²H = 32

$\rightarrow$ H = $\dfrac{32}{L^2}$

Surface Area A = LB + 2( L + B) H

A = L² + 2(L + L)($\dfrac{32}{L^2}$)

A= L² + $\dfrac{128}{L}$

$\dfrac{dA}{dL}=2L-\dfrac{128}{L^2}$

$2L-\dfrac{128}{L^2}=0$

$\rightarrow$ L³ = 64

$\rightarrow$ L = 4

$\dfrac{d^2A}{dL^2}=2+\dfrac{256}{L^3}$ ( + Ve)

Hence Area is minimum at L = 4

Surface Area = $L^2+\dfrac{128}{L^2}=4^2+\dfrac{128}{4}=16+32=48\;feet^2$

H = $\dfrac{32}{L^2}=\dfrac{32}{4^2}=2$

Dimension of Box = 4 * 4 * 2 feet

Answer:

Thus, 48 square feet is the least amount of material for its construction.