A rectangular box without a lid is to be made from 48 m2 of cardboard. Find the maximum volume of such a box. SOLUTION We let x, y, and z to be the length, width, and height, respectively, of the box in meters. Then we wish to maximize V

The maximum volume of such box is 32m^3

V = x×y×z = 32 m^3

Step-by-step explanation:

Given;

Total surface area S = 48m^2

Volume of a rectangular box V = length×width×height

V = xyz 1

Total surface area of a rectangular box without a lid is

S = xy + 2xz + 2yz = 48 2

To be able to maximize the volume, we need to reduce the number of variables.

Let assume the rectangular box has a square base,that means; length = width

x = y

Substituting y with x in equation 1 and 2;

V = x^2(z) 3

x^2 + 4xz = 48 4

Making z the subject of formula in equation 4

4xz = 48 - x^2

z = (48 - x^2)/4x 5

To be able to maximize V, we need to reduce the number of variables to 1, by substituting equation 5 into equation 3

V = x^2 × (48 - x^2)/4x

V = (48x - x^3)/4

differentiating V with respect to x;

V' = (48 - 3x^2)/4

At the maximum point V' = 0

V' = (48 - 3x^2)/4 = 0

Solving for x;

3x^2 = 48

x = √(48/3)

x = √(16)

x = 4

Since x = y

y = 4

From equation 5;

z = (48 - x^2)/4x

z = (48 - 4^2)/4(4)

z = 32/16

z = 2

The maximum volume can be derived by substituting x,y,z into equation 1;

V = xyz = 4×4×2 = 32 m^3

1/3 + 1/3 ≠ 2/6 because 1/3 alone = 2/6  

step by step explanation:

1/3 + 1/3 = 2/3       2/3=4/6