The volume of a box, is given by the function
[tex]V(x)=x(10-2x)(60-2x)=x(600-20x-120x+4 x^{2} )[/tex]
[tex]=4 x^{3}-140 x^{2} +600x[/tex]
The critical values, that is the points where the graph of the function takes its minimal and maximal values are found by solving V'(x)=0.
[tex]V'(x)=12 x^{2} -280x+600=4(3x^{2} -70x+150)=0[/tex]
we solve [tex]3x^{2} -70x+150=0[/tex]
a=3, b=-70, c=150
[tex]D= b^{2}-4ac= (-70)^{2}-4*3*150=4900-1800=3100
[/tex]
square root of 3100 is ≈ 55.68
[tex]x_1= \frac{-b+ \sqrt{D} }{2a}= \frac{70+ 55.68}{6}=20.95[/tex]
[tex]x_2= \frac{-b- \sqrt{D} }{2a}= \frac{70- 55.68}{6}=2.39[/tex]
x cannot be 20.95, because the side 10-2x would be negative, which makes no sense.
So x= 2.39
Answer: 2.39
