If an open box is made from a tin sheet 6 in. square by cutting out identical squares from each corner and bending up the resulting flaps, determine the dimensions of the largest box that can be made. (Round your answers to two decimal places.)

4in. square x 4in. square x 1in. square

Step-by-step explanation:

By cutting out identical squares (considering 'x') from each corner and bending up the resulting flaps

The dimensions we have will be (see attachment for the figure)

V= (6-2x) (6-2x) x

As, Volume'V' = length (l) x width(w) x height(h)

V= (6x- 2x²)(6-2x)

V= 36x - 12x²-12x²+ 4x³

V=4x³ - 24x²+ 36x

Next is to find dV/dx,therefore we find the derivative and and set it to zero for the maximum volume

dV/dx = 12x² - 48x + 36

setting it to zero

12x² - 48x + 36 =0

x² - 4x + 3=0

x² -3x -x + 3=0

x(x-3) -1(x-3)=0

Either : x-3=0=> x=3

OR : x-1 =0 => x=1

Now, notice that 'x' cannot be 3 , because if  we cut 3 inch squares out of the original square, there  will be nothing left!

Also,  the volume will be 0 then. That  is the minimum volume, 0, when we cut all the tin away.

So, x=1

Therefore,

height 'x' = 1in. square

length and width = (6-2x) => 4in. square

hello from mrbilldoesmath!

answer:

25x^2 + 90x + 84

discussion:

mn(x) = m(n(x)) = m( 5x + 9) =

(5x + 9)^2 + 3 =>

25x^2 + 90x + 9^2 + 3 =

25x^2 + 90x + 84                 (as 9^2 = 9*9 = 81)

which is the second choice

regards,  

mrb

p.s.   i'll be on vacation from friday, dec 22 to jan 2, 2019. have a great new year!