Answer to Question #110752 in Calculus for DertySol

Question #110752

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1. Use Lagrange Multiplier to determine the dimensions of a rectangular box, open at the top, having a volume of 32 cubic feet and requiring the least amount of material for its construction.


1
Expert's answer
2020-04-20T15:20:49-0400

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

Rectangular box, open at the op

Given : Volume=abc=32ft3Volume = abc=32ft^3

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

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


Thus, V(a,b,c)=abc32=0V(a,b,c)=abc-32=0

And S(a,b,c)=ab+2bc+2caS(a,b,c)=ab+2bc+2ca = minimum.

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

S=λV\triangledown S=\lambda \triangledown V

S=[SaSbSc]=[b+2ca+2c2a+2b];\triangledown S=\begin{bmatrix} S_a\\ S_b \\S_c \end{bmatrix}=\begin{bmatrix} b+2c\\ a+2c \\2a+2b \end{bmatrix}; V=[VaVbVc]=[bcacab]\triangledown V=\begin{bmatrix} V_a\\ V_b \\V_c \end{bmatrix}=\begin{bmatrix} bc\\ ac \\ab \end{bmatrix}

    [b+2ca+2c2a+2b]=λ[bcacab]\implies \begin{bmatrix} b+2c\\ a+2c \\2a+2b \end{bmatrix}=\lambda \begin{bmatrix} bc\\ ac \\ab \end{bmatrix}

    [b+2ca+2c2a+2b]=[λbcλacλab]\implies \begin{bmatrix} b+2c\\ a+2c \\2a+2b \end{bmatrix}= \begin{bmatrix} \lambda bc\\ \lambda ac \\\lambda ab \end{bmatrix}

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

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

    b=a\implies b=a

(a+2c)/ac=(2a+2b)/ab    (a+2c)/ac=(a+2c)/ac=(2a+2b)/ab \implies (a+2c)/ac=4a/a2=4/a    a/c+2=44a/a^2=4/a \implies a/c+2=4

    c=a/2\implies c=a/2


V=abc32=a(a)(a/2)32=0    a364=0V=abc-32=a(a)(a/2)-32=0 \implies a^3-64 =0

    a=4;b=4;c=2\implies a=4; b=4; c=2

    S=ab+2bc+2ca=16+16+16=48ft2\implies S= ab+2bc+2ca=16+16+16=48 ft^2


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


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