You are to manufacture a rectangular box with 3 dimensions x, y and z, and volume v=8000. Find the dimensions which minimize the surface area of this box.

20 by 20 by 20

Step-by-step explanation:

Let the total surface of the rectangular box be expressed as S = 2xy + 2yz + 2xz

x is the length of the box

y is the width and

z is the height of the box.

S = 2xy + 2yz + 2xz ... 1

Given the volume V = xyz = 8000 ... 2

From equation 2;

z = 8000/xy

Substituting into equation 1;

S = 2xy + 2y(8000/xy)+ 2x(8000/xy)

S = 2xy+16000/x+16000/y

Differentiating the resulting equation with respect to x and y will give;

dS/dx = 2y + (-16000x⁻²)

dS/dx = 2y - 16000/x²

Similarly,

dS/dy = 2x  + (-160000y⁻²)

dS/dy = 2x - 16000/y²

Note that at the turning point, ds/dx = 0 and ds/dy = 0, hence;

2y - 16000/x² = 0 and 2x - 16000/y² = 0

2y = 16000/x² and 2x = 16000/y²

2y = 16000/(8000/y²)²

2y = 16000×y⁴/64,000,000

2y = y⁴/4000

y³ = 8000

y =³√8000

y = 20

Given 2x = 16000/y²

2x = 16000/20²

2x = 16000/400

2x = 40

x = 20

Since Volume of the box is V = xyz

8000 = 20(20)z

8000 = 400z

z = 8000/400

z = 20

Hence, the dimensions which minimize the surface area of this box is 20 by 20 by 20.

the distance between point a and point b is 6 units