if we have a square base; then:
long=width=x
height =h
Area of this open rectangular box=x²+4(xh)
then:
x²+4xh=48
4xh=48-x²
h=(48-x²)/4x
let:
V(x,h)=volume of this open rectangular box
V(x,h)=x²h
V(x)=x²(48-x²)/4x
V(x)=x(48-x²)/4
V(x)=(48x-x³)/4
1)we have to find the first derivative of V(x):
V´(x)=(48-3x²)/4
2)We find out the values of "x", When V´(x)=0
(48-3x²)/4=0
48-3x²=0*4
48-3x²=0
-3x²=-48
x²=-48/-3
x²=16
x=⁺₊√16
We have two possible solutions:
x=-4; this is not solution, because the long or width can´t be negative.
x=4;it is possible solutions
3)we have to find the second derivative
V´´(x)=-6x/4=-3x/2
V´´(4)=-12/2=-6<0; then we have a maximum at x=4;
4) we find the value of "h", when x=4
h=(48-x²)/4x
h=(48-4²)/4(4)
h=(48-16)/16
h=32/16
h=2
Therefore:
long=width=4 ft
height=2 ft
V(x,y)=x²h=(4 ft)²(2ft)=32 ft³
Answer: the largest possible volume would be 32 ft³
