Posted by Sushmitha on Saturday, October 15, 2011 at 1:57pm.
The height of the box becomes x, and each side is reduced by 2x.
Therefore the volume is given by:
V = x ( 33-2x )( 33 - 2x )
because the original length and width are both 33.
How to find maximum? I don't know .
so how do i find the max volume?
Not much help from just algebra. If you can't use calculus, your best bet is to graph the function and approximate the value of y at the top of the hump.
1st of all I learned this in my calculus class. Now to the problem
now since you are going to cut squares (X)from a sheet of 33*33 inches. I am using capital 'X' instead of small
the volume would be
V = (33 - 2X)*(33 - 2X)* X
where one (33 - 2X) is length the other is width and 'X' is height
then use FOIL or in other words multiply it out
(4X^2 - 66X - 66X + 1089)* X
(4X^2 - 132X^2 + 1089)* X
therefore the final volume equation
V = (4X^3 - 132X^2 + 1089X)
now take the derivative of the volume equation
V' = (12X^2 - 264X + 1089)
now set the equation equal to zero
12X^2 - 264X + 1089 = 0
now you could have factored out 12 from the equation to make it easier but 1089 divided by 12 comes out to a decimal.
So I used the quadratic formula, so when you plug in the values, I couldn't put in the symbol for square root so just pretend / means square root and everything inside the curly brackets {} are inside the square root, then it would be:
-(-264) + or - /{(-264)^2 - 4*12*1089}
everything divided by (2*12)
264 + or - /{17424} all divided by 24
264 + or - 132 all divided by 24
now
264 + 132 divided by 24 gives you 16.5
and
264 - 132 divided by 24 gives you 5.5
therefore x = 5.5 and 16.5
now we take the second derivative to find the maximum value
so
V' = (12X^2 - 264X + 1089)
take the 2nd derivative of the equation above
V'' = (24X - 264)
now we plug in the 2 values of 'X' we got in this equation
(24*16.5) - 264 = 132
&
(24*5.5) - 264 = -132
since we want to maximize volume we would take 5.5 as the value of X because the answer is negative (the graph would be concave down, therefore maximum point)
we know have the value of 'X' so plug it into the dimensions that we started with
(33 - (2*5.5)) = 22
therefore
the dimensions that the box should be cut are 22 inches by 22 inches by 5.5 inches
*** if you would like to know the maximum volume of the box plug in 5.5 into the volume equation
V = (4X^3 - 132X^2 + 1089X)
= (4*5.5)^3 - (132*5.5)^2 + (1089*5.5)
= 2662 inches^3