Solution. \[D = \frac {\left |-2+0+1-8\right |}{\sqrt {1+1+1}} = \frac {9}{\sqrt 3} = 3\sqrt 3 \approx 5.196 .\] The Lagrange multiplier route gives the same answer with far more work; the formula already encodes the perpendicular from the point to the plane.

Solution. Let the box have edges \(2x\), \(2y\), \(2z\) with a corner on the sphere \(x^2+y^2+z^2 = a^2\), so \(V = 8xyz\). By Lagrange multipliers, \[8yz = 2\lambda x,\qquad 8xz = 2\lambda y,\qquad 8xy = 2\lambda z .\] Multiplying the first by \(x\), the second by \(y\) and the third by \(z\) gives \[8xyz = 2\lambda x^2 = 2\lambda y^2 = 2\lambda z^2 .\] Since the volume is non-zero, \(\lambda \neq 0\) and hence \(x^2 = y^2 = z^2\). With all three positive, \(x = y = z\), so the box is a cube. The constraint then gives \(x = \dfrac {a}{\sqrt 3}\) and \[V = \frac {8a^3}{3\sqrt 3}.\] This is the case \(a=b=c\) of the ellipsoid question on tutorial sheet 1, and it is worth seeing that the general answer specialises correctly.

Solution. \[f_x = 2x - \frac {2}{x^2y^2} = 0,\qquad f_y = 2y - \frac {4}{xy^3} = 0 .\] The first gives \(x^3y^2 = 1\) and the second \(xy^4 = 2\). dividing, \[\frac {x^3y^2}{xy^4} = \frac 12 \implies \frac {x^2}{y^2} = \frac 12 \implies y^2 = 2x^2 .\] Substituting into \(x^3y^2 = 1\) gives \(2x^5 = 1\), so \[x = 2^{-1/5},\qquad y = \pm 2^{3/10}.\] At both points the discriminant is \(D = 40 > 0\) and \(f_{xx} = 6 > 0\), so each is a local minimum, with value \[f = \frac {5\cdot 2^{3/5}}{2}\approx 3.79 .\] Note there are two of them, symmetric in \(y\), because \(y\) appears only through \(y^2\).

Solution. \[f_x = ye^{xy},\quad f_y = xe^{xy},\quad f_{xx} = y^2e^{xy},\quad f_{xy} = (1+xy)e^{xy},\quad f_{yy} = x^2e^{xy}.\] At \((1,2)\), where \(e^{xy} = e^2\), these are \[f = e^2,\quad f_x = 2e^2,\quad f_y = e^2,\quad f_{xx} = 4e^2,\quad f_{xy} = 3e^2,\quad f_{yy} = e^2 .\] With \(h = x-1\) and \(k = y-2\), \[e^{xy} \approx e^2\Big [1 + 2h + k + \frac 12\big (4h^2 + 6hk + k^2\big )\Big ] = e^2\Big [1 + 2h + k + 2h^2 + 3hk + \frac {k^2}{2}\Big ].\]