Section 14.8: Lagrange Multipliers

SOLUTION As in Example 14.7.6, we let \(x, y\), and \(z\) be the length, width, and height, respectively, of the box in meters. Then we wish to maximize \[ V = xyz \] subject to the constraint \[ g(x, y, z) = 2xz + 2yz + xy = 12 \] Using the method of Lagrange multipliers, we look for values of \(x, y, z\), and \(\lambda\) such that \(\nabla V = \lambda \nabla g\) and \(g(x, y, z) = 12\). This gives the equations \[ V_x = \lambda g_x \quad V_y = \lambda g_y \quad V_z = \lambda g_z \quad 2xz + 2yz + xy = 12 \] which become \[ yz = \lambda(2z + y) \] \[ xz = \lambda(2z + x) \] \[ xy = \lambda(2x + 2y) \] \[ 2xz + 2yz + xy = 12 \] There are no general rules for solving systems of equations. Sometimes some ingenuity is required. In the present example you might notice that if we multiply (2) by \(x\), (3) by \(y\), and (4) by \(z\), then the left sides of these equations will be identical. Doing this, we have \[ xyz = \lambda x(2z + y) \] \[ xyz = \lambda y(2z + x) \] \[ xyz = \lambda z(2x + 2y) \] We observe that \(\lambda \neq 0\) because \(\lambda = 0\) would imply \(yz = xz = xy = 0\) from (2), (3), and (4) and this would contradict (5). Therefore, from (6) and (7), we have \[ x(2z + y) = y(2z + x) \implies 2xz + xy = 2yz + xy \implies xz = yz \] But \(z \neq 0\) (since \(z = 0\) would give \(V = 0\)), so \(x = y\). From (7) and (8) we have \[ y(2z + x) = z(2x + 2y) \implies 2yz + xy = 2xz + 2yz \implies xy = 2xz \] Since \(x \neq 0\), we have \(y = 2z\). If we now put \(x = y = 2z\) in (5), we get \[ (2z)(2z) + 2(2z)z + 2(2z)z = 12 \implies 4z^2 + 4z^2 + 4z^2 = 12 \implies 12z^2 = 12 \] Since \(x, y\), and \(z\) are all positive, we therefore have \(z = 1\) and so \(x = 2\) and \(y = 2\). This agrees with our answer in Section 14.7.