Answer:
C(min) = 277.95 $
Container dimensions:
x = 2.822 m
y = 1.411 m
h = 3.52 m
Step-by-step explanation:
Let´s call x and y the sides of the rectangular base.
The surface area for a rectangular container is:
S = Area of the base (A₁) + 2 * area of a lateral side x (A₂) + 2 * area lateral y (A₃)
Area of the base is :
A₁ = x*y we assume, according to problem statement that
x = 2*y y = x/2
A₁ = x²/2
Area lateral on side x
A₂ = x*h ( h is the height of the box )
Area lateral on side y
A₃ = y*h ( h is the height of the box )
s = x²/2 + 2*x*h + 2*y*h
Cost = Cost of the base + cost of area lateral on x + cost of area lateral on y
C = 10*x²/2 + 8* 2*x*h + 8*2*y*h
C as function of x is:
The volume of the box is:
V(b) = 14 m³ = (x²/2)*h 28 = x²h h = 28/x²
C(x) = 10*x²/2 + 16*x*28/x² + 16*(x/2)*28/x²
C(x) = 5*x² + 448/x + 224/x
Taking derivatives on both sides of the equation we get:
C´(x) = 10*x - 448/x² - 224/x²
C´(x) = 0 10x - 448/x² - 224/x² = 0 ( 10*x³ - 448 - 224 )/x² = 0
10*x³ - 448 - 224 = 0 10*x³ = 224
x³ =22.4
x = ∛ 22.4
x = 2.822 m
y = x/2 = 1.411 m
h = 28/x² = 28 /7.96
h = 3.52 m
To find out if the container of such dimension is the cheapest container we look to the second derivative of C
C´´(x) = 10 + 224*2*x/x⁴
C´´(x) = 10 + 448/x³ is positive then C has a minimum for x = 2.82
And the cost of the container is:
C = 10*(x²/2) + 16*x*h + 16*y*h
C = 39.82 + 158.75 + 79.38
C = 277.95 $
