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astraxan [27]
3 years ago
5

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 fou

r corners and bending up the sides. Find the maximum volume that the box can have.
Mathematics
1 answer:
Burka [1]3 years ago
6 0

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.

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