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

solution:

the given function is ,

      =

the three roots of this quadratic function is , 0 ,8 and 10.all the roots are real and rational.

if a quadratic equation is of the form ,

sum of the roots = a + b +c= 0+8+10=18=  

product of roots taken two at a time = a b + b c+ c a = 0+80+0=80 =  

product of roots = a b c = 0=  

there are 11 doors in a hall starting from 0 to 11.the entry in the hall is permitted through gate number 0 and exit is allowed through only gate number 8 and 10.so, three roots(0,8,10) of quadratic function is given by entry and exit route of this hall.

answer: what in the world is this

step-by-step explanation: