an open-topped box can be made from a rectangular sheet of aluminum, with dimensions 40 cm by 25 cm, by cutting equal-sized squares from the four corners and folding up the sides.
Declare your variables and write a function to calculate the volume of a box that can be formed.
I figured out this part because the function would be f(x)=x(25-2x)(40-2x)
x being the height
25-2x being the width
40-2x being length
Then it asks what cut lengths to the nearest hundredth are acceptable if the volume of the box must be between 1512 and 2176cm^3.
So you would the write 1512<x(40-2x)(25-2x)<2176
how to you solve the inequality to get the x
PLEASE HELP THNX I HAVE A TEST TMR
Damon you answered this but they cant both equal zero its between those numbers
Look at my last reply to you
http://www.jiskha.com/display.cgi?id=1395791951#1395791951.1395793571
you need to solve two inequalities:
x(40-2x)(25-2x) > 1512
x(40-2x)(25-2x) < 2176
x(40-2x)(25-2x) = 4x^3 - 130x^2 + 1000x
So, you have to solve
4x^3 - 130x^2 + 1000x - 1512 > 0
4x^3 - 130x^2 + 1000x - 2176 < 0
a little synthetic division shows that the first has a root at x=2 and the second has a root at x=4
Now you are just left with two quadratics to solve, and that's easy
There are 3 intervals in the solution, but only two are feasible given positive dimensions.
The two curves are shown here:
http://www.wolframalpha.com/input/?i=plot+y%3D4x^3+-+130x^2+%2B+1000x+-+2176+and+y%3D4x^3+-+130x^2+%2B+1000x+-+1512
what do you mean by intervals
the solution set to an inequality is not just a number, but a whole range of numbers which satisfy the inequality.
Any x between 2 and 4 is a solution to this problem, as well as in the other intervals. Graphed on the number line, we have
http://www.wolframalpha.com/input/?i=solve+4x^3+-+130x^2+%2B+1000x+-+2176+%3C+0+and+4x^3+-+130x^2+%2B+1000x+-+1512+%3E+0
the plate is only 25 cm wide so the big x interval will not work
I got results in the middle of the first two intervals though