Question

From a thin piece of cardboard 30 in. by 30 in., square corners are cut out so that the sides can be folded up to make a box. What dimensions will yield a box of maximum volume? What is the maximum volume? Round to the nearest tenth, if necessary.

Answer 1

Answer:

Dimensions 5in x 10in x 10in will yeild a box with a max. volume of  500 cubic inches

Step-by-step explanation:

Volume =  height x length x width

considering 'x' as the length of the square corners that has been cut out from the cardboard and also, that is height of the cardboard box.

square corners are cut out so that the sides can be folded up to make a box, cardboard sides would reduce by 2x

therefore,

V = x  (30-2x) (30-2x) ---> eq(1)

V=( 30x - 2x²) (30-2x)

V= 900x- 60x² - 60x² + 4x³

V=  4x³ - 120 x²+ 900x

Taking derivative w.r.t 'x'

dV/dx = 12x² - 240 x +900

dV/dx = 4 (3x² - 60x +225)

For maximum dV/dx, make it equal zero

dV/dx = 0

so,  4 (3x² - 60x +225)=0

3x² - 60x+225=0   (taking 3 common)

x² - 20x + 75 =0

Solving this quadratic equation

x² - 15x -5x + 75 =0

x(x-15) - 5(x-15) =0

Either (x-15)=0

x=15

Or x-5=0

x = 5

if we substitute x=15 in eq(1), volume becomes zero.

therefore, x cannot be 15

When x= 5

eq(1)=>V = 5  (30-2(5)) (30-2(5))

V= 5 (10) (10)

V= 500 cubic inches,

Therefore, Dimensions 5in x 10in x 10in will yeild a box with a max. volume of  500 cubic inches