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let x he the height of the box... length = 18 - 2x width = 11 - 2x |dw:1433120436164:dw| the volume is \[V = x (18-2x)(11-2x)\] multiply it out
when you get an equation, then find the 1st derivative. let it equal zero and solve for x
\[V=x(18-2x)(11-2x) \] \[V=(18x-2x^2)(11x-2x^2)\] \[V=198x^2-36x^3-22x^3+4x^4\] \[V=198x^2-58x^3+4x^4\]
well you may want to check, as I think the equation is a cubic... V = x(198 - 58x + 4x^2) then distribute this V = 198x - 58x^2 + 4x^3 now find the 1st derivative
yes, it is cubic- and would it be 198-116x+12x^2?
now solve for x
x= 2.214 and 7.453 Now what do I do?
well you can't use 7.453... as the width is 11-2x is x is 7.453 the width is -4....which is impossible... so it must be the smaller choice...is probabbly the solution... just check by substituting
hope that makes sense... the alternative is to find the 2nd derivative... and test the solutions in the 2nd derivative to determine the max value
Yeah, once I plug it in it gives me 197.48 as the answer.. and that's the maximum volume right?
I haven't done the caclulations... you can check by graphing thecubic equation
Okay, I see how this worked out, thank you
I just graphed it and the smaller value gives the max volume
the max is (2.2, 197.5) hope it helped
yes, it did thanks