Question

Use langrange multiplier to determine the dimension of rectangular box open at the top, having a volume of 32ft^3 and requiring the least amount of material for it's construction

Answer 1

Let dimensions of the box be a, b and c feet respectively.

Rectangular box, open at the op

Given : $Volume = abc=32ft^3$

To find : least amount of material for its construction, which means minimum surface area.

Solution : $S=ab+2bc+2ca$ is the surface area of the required box.

Thus, $V(a,b,c)=abc-32=0$

And $S(a,b,c)=ab+2bc+2ca$ = minimum.

Using the Method of Lagrange's Multiplier, we get;

$\triangledown S=\lambda \triangledown V$

$\triangledown S=\begin{bmatrix} S_a\\ S_b \\S_c \end{bmatrix}=\begin{bmatrix} b+2c\\ a+2c \\2a+2b \end{bmatrix};$ $\triangledown V=\begin{bmatrix} V_a\\ V_b \\V_c \end{bmatrix}=\begin{bmatrix} bc\\ ac \\ab \end{bmatrix}$

$\implies \begin{bmatrix} b+2c\\ a+2c \\2a+2b \end{bmatrix}=\lambda \begin{bmatrix} bc\\ ac \\ab \end{bmatrix}$

$\implies \begin{bmatrix} b+2c\\ a+2c \\2a+2b \end{bmatrix}= \begin{bmatrix} \lambda bc\\ \lambda ac \\\lambda ab \end{bmatrix}$

$\implies \lambda = (b+2c)/bc=(a+2c)/ac=(2a+2b)/ab$

$(b+2c)/bc=(a+2c)/ac \implies 1+2c/b=1+2c/a$

$\implies b=a$

$(a+2c)/ac=(2a+2b)/ab \implies (a+2c)/ac=$$4a/a^2=4/a \implies a/c+2=4$

$\implies c=a/2$

$V=abc-32=a(a)(a/2)-32=0 \implies a^3-64 =0$

$\implies a=4; b=4; c=2$

$\implies S= ab+2bc+2ca=16+16+16=48 ft^2$

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