A company is manufacturing an open-top rectangular box. They have 30 cm by 16 cm sheets of material. The bins are made by cutting squares the same size from each corner of a sheet, bending up the sides, and sealing the corners. Create an equation relating the volume V of the box to the length of the corner cut out x. Identify the dimensions of the box that will have the maximum volume.

a. (30 - 2x)(16 - 2x)x. b. length 21.2 cm , width = 7.2 cm and height x = 4.4 cm

Step-by-step explanation:

a. Let x be the side of the squares to be cut from each corner. Since we have two corners on each side, the length of the resulting box from the 30 cm by 16 cm sheet is L = 30 - 2x. Its breadth is B = 16 - 2x. The height of the resulting box is x. So its volume V = LBx = (30 - 2x)(16 - 2x)x.

b. To find the maximum value of V, we differentiate V with respect to x and equate it to zero.

So, dV/dx = 0

d(30 - 2x)(16 - 2x)x/dx = 0

-2(16 - 2x)x + (-2)(30 - 2x)x + (30 - 2x)(16 - 2x) = 0

32x + 4x² - 60x - 4x² + 480 - 60x - 32x + 4x² = 0

4x²- 120x + 450 = 0

x²- 30x + 112.5 = 0

Using the quadratic formula, we find x. So, with a = 1, b = - 15 and c = 112.5,

x ≅ 25.61 or 4.4

V = (30 - 2x)(16 - 2x)x

Substituting the values of x into L and B, we have )x

L = (30 - 2x) = (30 - 2(25.61)) = 30 - 51.22 = -21.22

B = (16 - 2x) = (16 - 2(25.61)) = 16 - 51.22 = -35.22

Since L and B cannot be negative, we use the other value for x = 4.4, So

L = (30 - 2x) = (30 - 2(4.4)) = 30 - 8.8 = 21.2

B = (16 - 2x) = (16 - 2(4.4)) = 16 - 8.8 = 7.2

So V = LBx = 21.2 × 7.2 × 4.4 = 671.62 cm³

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15