A rectangular box without lid is to be made from 12 m² of cardboard. Find the maximum volume of such box.

From the information given, let assume x, y, z are the length, width, height of the box respectively( in meters).

To find max. Volume (V) = xyz by applying Lagrange multiplier;

Subject to constraint

g(x,y,z) = z (xy+yz)+(zx) – 12

Now; F = f(x,y,z) + λg(x,y,z)

This implies that:

xyz + [2xy + 2yz +zx -12]

∴

F_x = yz + λ[2y + z]

F_y = xz + λ[2x +2z]

F_z = xy + λ[2y + x]

But F_x = F_y = F_z = 0

∴

yz + λ[2y + z] = λz + λ[2x + 2z] = xy + λ [2y +x]

⇒ x[2y +z] = y[2x+2z] = z[2y + x]

x[2y + z] = y[2x + 2z]

⇒ x = 2y

y[2x + 2z] = z[2y + x]

2y = z

and

2[xy + yz ] + zx = 12

⇒ 2[2y² + 2y²] + 4y² = 12

⇒ 12y² = 12

⇒ y² = 12/12

⇒ y² = 1

⇒ y = √1

⇒ y = 1

∴

x = 2y

x = 2(1)

x = 2

2y = z

2(1) = z

2 = z

z = 2

Hence, the max. Volume = (2)(1)(2) = 4m³