Question

An open box is to be made from a 3 ft by 8 ft rectangular piece of sheet metal by cutting out squares of equal size from the four corners and bending up the sides. Find the maximum volume that the box can have.

Answer 1

Answer:

Maximum volume of the box will be 7.41 cubic feet.

Step-by-step explanation:

Open box has been made from a metal sheet measuring 3 ft and 8 ft.

Let four square pieces were removed from the four corners with one side measurement x ft.

Volume of the open box = Length × width × height

Length of the box = (3 - 2x) ft

Width of the box = (8 - 2x) ft

Height of the box = x ft

Volume of the box = (3 - 2x)(8 - 2x)x

V = (24 - 6x - 16x + 4x²)x

V = 24x - 22x² + 4x³

Now we take the derivative of V with respect to x and equate the derivative to zero,

$\frac{d}{dx}(V)=\frac{d}{dx}(24x - 22x^{2}+4x^{3})$

V' = 24 - 44x + 12x²

V' = 0

12x² - 44x + 24 = 0

3x² - 11x + 6 = 0

3x² - 9x - 2x + 6 = 0

3x(x - 3) - 2(x - 3) = 0

(3x - 2)(x - 3) = 0

(3x - 2) = 0

and (x - 3) = 0

Therefore, x = 3, $\frac{2}{3}$

For x = 0.67

Length of the box = (3 - 2x) = 3 - 1.34

                              = 1.66 ft

Width of the box = (8 - 2x) = 8 - 1.34

                            = 6.66 ft

Volume of the box = 0.67 × 1.66 × 6.66

V = 7.41 cubic feet.

Similarly, for x = 3,

Length of the box = (3 - 2\times 3) = -3

which is negative but the length of the box can not be negative.

Therefore, maximum volume of the box will be 7.41 cubic feet.