Asked by Anonymous
sandy is making a closed rectangular jewwellery box with a square base from two different woods . the wood for top and bottom costs $20/m^2. the wood for the side costs $30/m^2 . find dimensions that minimize cost of wood for a volume 4000cm^3?
Answers
Answered by
Reiny
let the base be x m by x m
let the height be y m
1 m^3 = 100^3 cm^3 = 1 000 000 cm^3
V= 4000 cm^3 = .004 m^3
V = x^2y
.004 = x^2y
y = .004/x^2
cost = 20(2x^2) + 30(4xy)
= 40x^2 + 120x(.004/x^2)
= 40x^2 + .048/x
d(cost)/dx = 80x - .048/x^2 = 0 for max/min of cost
80x = .048/x^2
x^3 = .0006
x = .0843 m , x = 8.43 cm
y = .5623 m , y = 56.23 cm
check my arithmetic, looks like a weird shape for a jewelry box.
check:
if x= .0843, y = .5623, then V = .004
cost = 40(.0843)^2 + .048/.0843 = .8536
try an x value slightly higher
let x = .085
cost = 40(.085)^2 + .048/.085 = .8537 , which is higher
try an x value slightly lower
let x = .084
cost = 40(.084)^2 + .048/.084 = .85366 which is also higher
let the height be y m
1 m^3 = 100^3 cm^3 = 1 000 000 cm^3
V= 4000 cm^3 = .004 m^3
V = x^2y
.004 = x^2y
y = .004/x^2
cost = 20(2x^2) + 30(4xy)
= 40x^2 + 120x(.004/x^2)
= 40x^2 + .048/x
d(cost)/dx = 80x - .048/x^2 = 0 for max/min of cost
80x = .048/x^2
x^3 = .0006
x = .0843 m , x = 8.43 cm
y = .5623 m , y = 56.23 cm
check my arithmetic, looks like a weird shape for a jewelry box.
check:
if x= .0843, y = .5623, then V = .004
cost = 40(.0843)^2 + .048/.0843 = .8536
try an x value slightly higher
let x = .085
cost = 40(.085)^2 + .048/.085 = .8537 , which is higher
try an x value slightly lower
let x = .084
cost = 40(.084)^2 + .048/.084 = .85366 which is also higher
Answered by
swag
lele
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