8.1 Functions of Several Variables

Problem 8.1.1. Find the minimum distance between the point \((2,0,1)\) and the plane \(-x+y+z = 8\).

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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.

Problem 8.1.2. Show that the rectangular box of greatest volume inscribed in a sphere is a cube.

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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 problem among the practice problems for functions of several variables, and it is worth seeing that the general answer specialises correctly.

Problem 8.1.3. Find and classify the stationary points of \(f(x,y) = x^2+y^2+\dfrac {2}{xy^2}\).

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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 \(y^2\).

Problem 8.1.4. Derive the Taylor series of \(f(x,y) = e^{xy}\) about \((1,2)\) up to second-order terms.

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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 \(k = y-2\), \[e^{xy} \approx e^2\left [1 + 2h + k + \frac 12(4h^2 + 6hk + k^2)\right ] = e^2\left [1 + 2h + k + 2h^2 + 3hk + \frac {k^2}{2}\right ].\]

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