Answer:
4m³
Step-by-step explanation:
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
