Question

A worker is to construct an open rectangular box with a square base and a volume of 196196 ft cubedft3. If material for the bottom costs ​$88​/ft squaredft2 and material for the sides costs ​$77​/ft squaredft2​, what dimensions will result in the least expensive​ box? What is the minimum​ cost? Let x be the length of one side of the base of the box. Express the total cost of the​ materials, C, in terms of x.

Answer 1

Answer:

Therefore the dimensions of the box is 7 ft by 7 ft by 4 ft.

Therefore the minimum cost to manufacture the box is $12936.

Step-by-step explanation:

Given the volume of the open box is 196 cube ft.

Let the one side of the base of the box be x.

Since the base of the box is square in shape.

The other side of the base is also x.

The area of the base of the box = x² square ft

Let the height of the box be h.

The volume of the box is = Area of the base× height

                                          =(x²×h) cube ft.

                                           =x²h cube ft.

According to the problem,

x²h=196

$\Rightarrow h=\frac{196}{x^2}$

The material cost for the bottom is $88 per square ft and the material cost for the sides $77  per square ft.

The material cost for the bottom of the box = $(88×x²)

                                                                         =$88x²

The area sides = 2(length+width)height

                       =2(x+x)h

                      =4xh square ft.

The material cost for the sides of the box =$(77×4xh)

                                                                    =$308 xh

Total cost to make the box is =$(88x²+308 xh)

∴C=88x²+308 xh    [ C is in dollar]

Now putting $h=\frac{196}{x^2}$

$C=88x^2+308x\times \frac{196}{x^2}$

$\Rightarrow C=88x^2+ \frac{60368}{x}$

Differentiating with respect to x

$C'=176x- \frac{60368}{x^2}$

Again differentiating with respect to x

$C''=176+ \frac{120736}{x^3}$

Now to find the minimum cost, we set C'=0

$\therefore 176x- \frac{60368}{x^2}=0$

$\Rightarrow 176x= \frac{60368}{x^2}$

$\Rightarrow x^3= \frac{60368}{176}$

$\Rightarrow x^3=343$

$\Rightarrow x=7$

Now, $C''|_{x=7}=176+ \frac{120736}{7^3}>0$

Therefore at x=7, the manufacturing cost will be minimum.

The height of the box is $h=\frac{196}{7^2}$ = 4ft

Therefore the dimensions of the box is 7 ft by 7 ft by 4 ft.

The manufacturing cost is $C=88.7^2+ \frac{60368}{7}$   [ putting x=7]

                                             =$12936

Therefore the minimum cost to manufacture the box is $12936.

Answer 2

The cost function in terms of x is: $\mathbf{C = 88x^2 + \frac{60368}{x}}$; the dimension that minimizes the cost is 7 ft by 7ft by 4 ft, and the minimum cost is $12936

The length of the base is x and the height is y.

So, the volume of the box is:

$\mathbf{V = x^2y}$

Substitute 196 for volume

$\mathbf{x^2y = 196}$

Make y the subject

$\mathbf{y = \frac{196}{x^2}}$

The surface area of the box is:

$\mathbf{S = x^2 + 4xy}$

The base costs $88 while the sides cost $77 per square feet.

So, the cost function is

$\mathbf{C = 88x^2 + 77 \times 4xy}$

$\mathbf{C = 88x^2 + 308xy}$

Substitute $\mathbf{y = \frac{196}{x^2}}$

$\mathbf{C = 88x^2 + 308x \times \frac{196}{x^2}}$

$\mathbf{C = 88x^2 + 308 \times \frac{196}{x}}$

$\mathbf{C = 88x^2 + \frac{60368}{x}}$

So, the cost function in terms of x is: $\mathbf{C = 88x^2 + \frac{60368}{x}}$

Differentiate

$\mathbf{C' = 176x - \frac{60368}{x^2}}$

Set to 0

$\mathbf{176x - \frac{60368}{x^2} = 0}$

Multiply through by x^2

$\mathbf{176x^3 - 60368 = 0}$

Add 60368 to both sides

$\mathbf{176x^3 = 60368}$

Solve for x^3

$\mathbf{x^3 = 343}$

Take cube roots

$\mathbf{x = 7}$

Substitute 7 for x in $\mathbf{y = \frac{196}{x^2}}$

$\mathbf{y = \frac{196}{7^2}}$

$\mathbf{y = \frac{196}{49}}$

$\mathbf{y = 4}$

So, the dimension that minimizes the cost is 7 ft by 7ft by 4 ft

Substitute 7 for x in $\mathbf{C = 88x^2 + \frac{60368}{x}}$

$\mathbf{C = 88 \times 7^2 + \frac{60368}{7}}$

$\mathbf{C = 12936}$

Hence, the minimum cost is $12936

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