Answer:
The primary equation, which is the surface area, we're trying minimize is
S(x,y) = (2x² + 6xy)
The constraint equation is
C(x,y) = x²y - 18
The width of the box that minimizes the surface area of the box is x = 3 ft.
The corresponding height that minimizes the surface area of the box is y = 2 ft.
Step-by-step explanation:
Width of the box = x
Height of the box = y
Length of the box = z = 2x
Volume of the box = 36 ft³
xyz = 36
2x²y = 36
x²y = 18
Constraint equation = x²y - 18
Surface area of a regular box
S(x, y, z) = 2(xy + xz + yz)
But this box is open at the top
S(x,y,z) = (2xy + xz + 2yz)
And length, z, is twice as much as the width, x
S(x, y) = (2xy + 2x² + 4xy) = (2x² + 6xy)
The primary equation, which is the surface area, we're trying minimize is
S(x,y) = (2x² + 6xy)
The constraint equation is
C(x,y) = x²y - 18
To now find the width of the box that minimizes the surface area, we use the substitution method.
From the constraint, we know that
x²y - 18 = 0
x²y = 18
y = (18/x²) (eqn 1)
We then substitute this into the primary function
S(x,y) = (2x² + 6xy)
S(x) = 2x² + 6x (18/x²) = 2x² + (108/x)
To minimize the primary function now,
At minimum surface area, (dS/dx) = 0 and (d²S/dx²) > 0
S(x) = 2x² + (108/x)
(dS/dx) = 4x - (108/x²)
4x - (108/x²) = 0
4x = 108/x²
4x³ = 108
x³ = 27
x = ∛27 = 3 ft.
(d²S/dx²) = 4 + (216/x³) = 4 + (216/27) = 12 > 0 (it is indeed the value of x that corresponds to a minimum point!)
y = (18/x²)
y = (18/9) = 2 ft.
Hence, the width of the box that minimizes its surface area is x = 3 ft.
Hope this Helps!!!
