math...minimum volume help!?

You don't mention if the box is open at the top. I will assume it is not.

Let the square base be of side x and the height of the box be h.

Take 10L to be 10000 cm³

Volume = x²h = 10000

so

h = 10000 / x²

Total surface area = area of both ends + area of the four sides.

S = 2x² + 4xh

S = 2x² + 4x(10000 / x²)

S = 2x² + 40000 / x

dS/dx = 4x - 40000 / x² = 0

<=> 4x = 40000 / x²

<=> x³ = 10000

Giving x = ³√(10000)

x = 10(³√10)

x = 21.54

x = 22 cm

h = 10000 / x²

= 10000 / (22)²

h = 20.66

h = 21 cm

Box has square base of 22 cm and height 21 cm.

Surface area S = 2x² + 40000 / x

= 2(22)² + 40000 / 22

= 2787

Minimum surface area is 2787 cm²

To check for minimum value you show that d²S/dx² > 0 for the value of x you have calculated (2nd derivative test).

Note : if the box is open top then S = x² + 40000 / x

and you would proceed along similar lines to the above.

Erm, firstly you have to find the area of the square based prism, then use equate dy/dx=0 then find the value given for its stationary value and equate back to area needed

Eg if equation for the area needed is A=x^3+x^2+x

then dA/dx=3x^2+2x+1

then for stationary value dA/dx=0 which is 3x^2+2x+1=0

Finally solve for x and equate back to the equation of area