Question 987179
The 20cm side would become {{{20-2x}}} long.
The 60cm side would become {{{60-2x}}} long.
The total volume of the box would be {{{V=(20-2x)(60-2x)x}}}
{{{V=4x^3-160x^2+1200x}}}
To find the maximum volume, take the derivative and set it equal to zero.
{{{dV/dx=12x^2-320x+1200}}}
{{{3x^2-80x+300=0}}}
{{{3(x^2-(80/3)x)+300=0}}}
{{{3(x^2-(80/3)x+(40/3)^2)+300=3(40/3)^2}}}
{{{3(x-40/3)^2=3(1600/9)-300}}}
{{{3(x-40/3)^2=1600/3-900/3}}}
{{{3(x-40/3)^2=700/3}}}
{{{(x-40/3)^2=700/9}}}
{{{x-40/3=0 +- (10/3)sqrt(7)}}}
{{{x=40/3 +- (10/3)sqrt(7)}}}
Of the two values, only {{{x=40/3-(10/3)sqrt(7)}}} is in the range of x which can only have values between {{{0}}} and {{{10}}} since {{{20-2x}}} must remain positive.
So then the maximum volume that the box can have is,
{{{V[max]=(56000sqrt(7)-80000)/27}}}