The base of a rectangular box is to be twice as long as it i

The base of a rectangular box is to be twice as long as it is wide. The volume of the box is 256 cubic inches. The material for the top costs $0.10 per square inch and the material for the sides and bottom costs $0.05 per square inch. Find the dimensions that will make the cost a minimum. length in width in height in

Solution

let length of base =x , width =y , height of box =z

given length is twice the width

=>x =2y

volume of box V =xyz

given volume =256

xyz =256

2y yz =256

z =128_/y²

total surface area of the box ,A=(xy +xy)+2(yz+xz)

given top costs 0.1 $ per sq in , sides and bottom cost 0.05 $ per sq in

total cost of box, C=(0.1xy +0.05xy)+2(0.05yz+0.05xz)

C=0.15xy +0.1yz+0.1xz

C=(0.15*2y*y) +(0.1yz)+(0.1*2y*z)

C=(0.3y²) +(0.1yz)+(0.2yz)

C=(0.3y²) +(0.3yz)

C=(0.3y²) +(0.3y*128_/y²)

C=(0.3y²) +(38.4_/y)

dor minimum cost dC/dy =0 ,d²C/dy² >0

dC/dy =(0.6y) -(38.4_/y²) ,d²C/dy² =(0.6) +(76.8_/y³)

(0.6y) -(38.4_/y²) =0

(0.6y³) =38.4

y³=64

y =4

x =2y

x =2*4

x=8

z =128_/y²

z =128_/4²

z=8

d²C/dy² =(0.6) +(76.8_/4³) >0

dimensions that will make cost a minimum are

length = 8 in

width =4 in

height =8 in