A company wishes to manufacture a box with a volume of 36 cubic feet that is open on top and is twice as long as it is wide. Find the width of the box that can be produced using the minimum amount of material. Write the two equations (Primary Function and Constraint) that are used to solve this problem. Let SS be the surface area of the box, VV its volume, xx the width and yy the height of the box. Group of answer choices S

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!!!

4096 bacteria

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