an open top box with a square base and a volume of 4m^3 is to be constructed what should the dimensions of the box be to minimize the surface area of the box?

Dimensions of the box:

x =  1,26  ( the side of the square base )

h = 2,52   the heigh of the cube

Step-by-step explanation:

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

The surface area of such cube is:

Area of the base   A(b)  = x²

Lateral area is  4*x*h

A(c) = x²  +  4*x*h

Now   x² * h  = v = 4 m³       ⇒  h = 4/x²   and

A(x)   =  x²   +  4/x

Tacking derivatives on both side of the equation:

A´(x)  =  2*x - 4/x²

A´(x) = 0       2*x  - 4/x²  =  0        ⇒   2*x³  -  4   = 0

x³  = 2

x  = 1,26 m

And  h = 4/(1,26)²        ⇒     h  =  2,52 m

How do we know the value of  x  =  1,26 is for a minimum value of A(x)

We find the second derivative of  A(x)

A´´(x)  =  2  -  (-4*2*x/x⁴)

A´´(x)  =  2  + 8/x³       is positive  A´´(x)  >  0

Then function A(x) has a minimum value at x = 1,26

hello from mrbilldoesmath!

answer:

a(n) - a(n-1) -3

discussion:

this ones a bit interesting. given a(n) = 5 - 3n consider this:

a(n-1) = 5 - 3(n-1)

          = (5 - 3n) + 3             => but a(n) = 5 - 3n, the first term

          = a(n)   + 3                 => subtract 3 from both sides

a(n-1) -3 = a(n)  

or

a(n) - a(n-1) -3

you,

mrb

the answer is a

step-by-step explanation: