Proof. Since \(\nabla g(x_0,y_0)\neq \mathbf 0\), the implicit function theorem (the subsection of that name) shows that near \((x_0,y_0)\) the curve \(g = k\) is the path of a differentiable curve \(\mathbf r(t)\) with \(\mathbf r(0) = (x_0,y_0)\) and \(\mathbf r'(0)\neq \mathbf 0\). The function \(t\mapsto f(\mathbf r(t))\) has a local extremum at \(t = 0\), so its derivative \(\nabla f\cdot \mathbf r'(0)\) is zero there. Also \(g(\mathbf r(t)) = k\) for all \(t\), so \(\nabla g\cdot \mathbf r'(0) = 0\). In the plane, both \(\nabla f\) and \(\nabla g\) are therefore perpendicular to the same non-zero vector \(\mathbf r'(0)\), hence parallel, and since \(\nabla g\neq \mathbf 0\), \(\nabla f = \lambda \nabla g\). On a surface the same reasoning applies to every curve through \((x_0,y_0,z_0)\) in the surface, so \(\nabla f\) is perpendicular to the whole tangent plane, as \(\nabla g\) is. □

Solution. With \(g = x - y + z\) and \(h = x^2 + y^2\), \[\nabla f = \langle 1, 2, 3\rangle ,\qquad \nabla g = \langle 1, -1, 1\rangle ,\qquad \nabla h = \langle 2x, 2y, 0\rangle ,\] so the equations are \begin {align} 1 & = \lambda + 2\mu x, \label {eq:tc1}\\ 2 & = -\lambda + 2\mu y, \label {eq:tc2}\\ 3 & = \lambda , \label {eq:tc3}\\ x - y + z & = 1, \label {eq:tc4}\\ x^2 + y^2 & = 1 . \label {eq:tc5} \end {align}

By (??), \(\lambda = 3\), and then (??) and (??) give \[2\mu x = -2,\qquad 2\mu y = 5,\qquad \text {so}\qquad x = -\frac {1}{\mu },\quad y = \frac {5}{2\mu }.\] Substituting into (??), \[\frac {1}{\mu ^2}+ \frac {25}{4\mu ^2} = 1 ,\qquad \mu ^2 = \frac {29}{4},\qquad \mu = \pm \frac {\sqrt {29}}{2},\] so \(x = \mp \frac {2}{\sqrt {29}}\) and \(y = \pm \frac {5}{\sqrt {29}}\), with the signs taken together. From (??), \(z = 1 - x + y = 1 \pm \frac {7}{\sqrt {29}}\). The two candidates are \[\left (-\frac {2}{\sqrt {29}}, \frac {5}{\sqrt {29}}, 1 + \frac {7}{\sqrt {29}}\right ) \quad \text {and}\quad \left (\frac {2}{\sqrt {29}}, -\frac {5}{\sqrt {29}}, 1 - \frac {7}{\sqrt {29}}\right ),\] where \[f = 3 \pm \frac {-2 + 10 + 21}{\sqrt {29}} = 3 \pm \frac {29}{\sqrt {29}} = 3\pm \sqrt {29} .\] The maximum value is \(3 + \sqrt {29}\approx 8.39\), and the minimum is \(3 - \sqrt {29}\).

Solution. Let the base be \(x\) by \(y\) and the height \(z\), and measure cost in units of the price of the top and sides per square metre. The cost is \(C = 2xy + xy + 2xz + 2yz = 3xy + 2xz + 2yz\), to be minimised subject to \(xyz = 32\). The Lagrange equations are \[3y + 2z = \lambda yz,\qquad 3x + 2z = \lambda xz,\qquad 2x + 2y = \lambda xy .\] Multiplying the first by \(x\) and the second by \(y\) and subtracting gives \(2z(x - y) = 0\), so \(x = y\). Multiplying the second by \(y\) and the third by \(z\) and subtracting gives \(3xy - 2yz = 0\) (using \(x = y\)), so \(z = \frac 32 x\). Then \(xyz = \frac 32 x^3 = 32\), so \(x^3 = \frac {64}{3}\) and \[x = y = \frac {4}{\sqrt [3]{3}}\approx 2.77\ \text {m},\qquad z = \frac 32 x\approx 4.16\ \text {m}.\] The more expensive base makes the box taller than a cube.

Solution. The equations \(yz = \lambda \), \(xz = \lambda \), \(xy = \lambda \) give \(yz = xz = xy\), and since the variables are positive, \(x = y = z\). The constraint then gives \(x = y = z = 4\) and the maximum \(f = 64\). (The maximum exists because \(f\) is continuous on the closed bounded triangle \(x,y,z\geq 0\), \(x+y+z = 12\) and vanishes on its edges.) Replacing \(12\) by any sum \(s\) gives the maximum \(\left (\frac {s}{3}\right )^3\), so for positive numbers with sum \(s\), \[xyz\leq \left (\frac {x+y+z}{3}\right )^3,\qquad \text {that is}\qquad \sqrt [3]{xyz} \leq \frac {x+y+z}{3},\] with equality only when \(x = y = z\).