i need a box with a square base and an open top that cna hold 500 in^3. what is the least amount of material i can use to build this box

M(x)  =  452,56 in

Step-by-step explanation:

The volume of the open box is  500 in³

V = Area of the base times height

V(b) =  x² * h               where x is the side of the square and h the heigh

Then     500 =  x²*h

Total material to use is: material of the base + material of 4 sides

material of the base is  x²

material of one side is x*h  we have 4 sides then  4*x*h

Total material M(b)

M(b)  =  x²  +  4*x*h

And    as    h  =  500/ x²

M(x)  =  x²  +  4* x* 500/x²

M(x)  =  x²  +  2000/x

Tacking derivatives on both sides of the equation

M´(x)  =  2*x   - 200/x²

M´(x)  =  0         2*x  -  200/x²  = 0

x³ -  100  =  0

x³   =  100

x   =  4,64 in

And By substitution     h  =  500/x²

h  =  500/(4,64)²           h  =  23,22  in

How do we know that x = 4,64 make V(x) minimum

we get the second derivative

M´´(x)  =  2  +  200*2x/ x⁴    =  2  +  400/x³

M´´(x) is always positive

M´´(x) > 0     then M(x) has a minimum for  x= 4,64 in

The least amount of material is:

M(x)  =  x²    +    2000/x

M(x)  =  (4,64)²  *  2000/4,64

M(x)  = 21,53  +  431,03

M(x)  =  452,56 in

step-by-step explanation:

you can add the two numbers and divide the result by 2. or you can find the gap between the two numbers and add half of this to the smaller number.

the answer is (-8,-5)

step-by-step explanation:

the endpoint to midpoint is half the length. the length is (-2,-1) so when you are at the midpoint just add it to it to find your answer -6+(-2)= -8 and -4+(-1)=-5 so (-8,-5)