By cutting away identical squares from each corner of a rectangular piece of cardboard and folding up the resulting flaps, an open box may be made. If the cardboard is 16 in. long and 6 in. wide, find the dimensions (in inches) of the box that will yield the maximum volume. (Round your answers to two decimal places if necessary.)

x=3.00 or x= 8.00

Explanation:

Let us denote side of the cut-out square as x inches

Then base of box = 16-2x by 6-2x

The volume = V = (16-2x)(6-2x)

= 96-3x-12x+4x

dV/dx = 96-44x+4x^2

= 0 for a max V

divide by 4

X^2 -11x +24=0

Using quadratic formula to solve the equation

X= (11± √11^2-(4×1×24)/2

(11± √25)/2

x= 3 or 8inches

From the equation of volume = V = x(16-2x)(6-2x)

Substitute

let x = 3 , V = 3(16-6)(6-6) = 0

let x = 8 , V = 8(16-16)(6-16) = 0

Therefore, the maximum volume exist at both point x=3 or x= 8

x = -4

step-by-step explanation:

- 20 - 4(x + 3) = 4x

=> -20 - 4x - 12 = 4x

=> - 32 -4x = 4x

=> -32 = 4x + 4x

=> - 32 = 8x

=> -32/8 = x

=> -4 = x