An open square box of volume 514,500 cubic inches has a square base and five partitions inside that are parallel to two sides. Find the dimensions of the box for which the amount of material needed to construct the box is as small as possible. Show all your work

x  side of the square base   equal to 127,20 in

h height of the cube             equal to  127,20 in

Step-by-step explanation:

Volume of cube Vc is   514500 in³

Let call x side of the square base, then

Area of the base  Ab = x²

Area of (4) lateral sides  Al  =  4 * x*h     h height of the cube

Area of (4) inside partitions Ap = 4*x*h

And Vc =  x²* h    ⇒  514500 = x²*h       ⇒  h   =  514500/ x²

Total area;  At = Ab + 4 Al + 4 Ap

as Al = Ap

At = Ab + 8 Al

At = x²  +  8*x*h               and   h = 2058000

Then area as function of x

A(x) = x² + 8*x* 514500/x²   ⇒        A(x) = x² +  4116000/x

Taking derivative on both sides of the equation

A´(x) = 2x - 4116000/x²       ⇒     A´(x) = 0

2x  - 4116000/x²  = 0

2*x³ - 4116000 = 0

x³  -  2058000 = 0

x³ = 2058000

x  = 127,20 in

And   h  =  2058000/x²        ⇒  h = 2058000/ 16179,84

h = 127,20 in

if this is really you s much. i only have 3 points left and i have 2 questions to ask

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: