Question

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?

Answer 1

Dimensions of the box:

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

h = 2,52 the heigh of the cube

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