The volume of a box with a square base and open top:
V = Ab ∙ h
Ab = Area of base = x²
V = Ab ∙ h = x² ∙ h
x² ∙ h = 32,000
Divide both sides by x²
h = 32,000 / x²
The surface area of the box = Area of base + 4 Area of rectangle
As = x² + 4 Ar
Ar = Area of rectangle = x ∙ h
As = x² + 4 ∙ x ∙ h
As = x² + 4 ∙ x ∙ 32,000 / x²
dAs / dx = As'₍ x₎ = ( x² + 4 ∙ x ∙ 32,000 / x² )' = 2 ( x³ - 64,000 ) / x²
If
f ' (x) = 0
function has a local maximum or a local minimum
As'₍ x₎ = 0
2 ( x³ - 64,000 ) / x² = 0
Multiply both sides by x² / 2
x³ - 64,000 = 0
Add 64,000 to both sides
x³ = 64,000
x = ∛64,000 = 40
x = 40 cm
The second derivative test:
When
f ' ₍x₀₎ = 0
then
if
f " ₍x₀₎ > 0, then f has a local minimum at x₀
if
f " ₍x₀₎ < 0, then f has a local maximum at x₀
In this case:
As"₍ x₎ = ( As'₍ x₎ )' = [ 2 ( x³ - 64,000 ) / x² ] ' = 256,000 / x³ + 2
for x = 40
As" = 256,000 / x³ + 2 = 256,000 / 40³ + 2 = 6 > 0
So the surface area of the box has a local minimum for x = 40 cm
Dimension of the box that minimizes the amount of material used:
x = 40 cm
h = 32,000 / x² = 32,000 / 40² = 20 cm
A box with a square base and open top just have volume of 32,000 cm^3 . find the dimension of the box that minimizes the amount of material used.
1 answer