Thread: maximum volume of a box

I have the volume equation as

V=510x-1/2x^3

then dv/dx = 0 therefore 510-3/2x^2=0

solving for x gives x= + or - (sqrt340)

(d^2v)/(dx^2)= -3x

when x = +(sqrt340)
(d^2v)/(dx^2)= negative therefore maximum

V=510(sqrt340)-0.5(sqrt340)^3 = 6269 m^3

I am not sure about the last part is it just a case of substituting the max x value obtained into the original volume formula?

Originally Posted by jlaing

I have the volume equation as

V=510x-1/2x^3

then dv/dx = 0 therefore 510-3/2x^2=0

solving for x gives x= + or - (sqrt340)

(d^2v)/(dx^2)= -3x

when x = +(sqrt340)
(d^2v)/(dx^2)= negative therefore maximum

V=510(sqrt340)-0.5(sqrt340)^3 = 6269 m^3

I am not sure about the last part is it just a case of substituting the max x value obtained into the original volume formula?

to calculate the maximum volume, yes.