Math21a, Fall 2000 Course Head: Prof. Clifford H. Taubes |

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Answers to selected problems in the Handout on 3-variable Lagrange multipliers: |

5) | The maximum, 14, occurs where x = 7 and y = 0. The minumum, -14, occurs where x = -7 and y = 0. |

6) | The problem is to minimize f(x,y) = 60 x + 100 y subject to the constraint
500 x^{0.04} y ^{0.08} = 10^{4}.
The minimum, 2400 (5/6)^{4/5} occurs at x = 20^{(5/6)}(5/6)^{2/3} and y = 24^{(5/6)}(5/6)^{-1/3}. |

7) | Denote the radius of the cylinder by r and the height by h. You are being asked to minimize
(r^{2} + 2 h r) subject to the constraint r^{2} h = 50.
The minimum occurs where r = (50/)^{1/3} and
h = (50/)^{1/3}. |

8) | The point has coordinates (101)^{-1/2} (8, 27, 4). One way to find this point is to realize that the
gradient of x^{2}/4 + y^{2}/9 + z^{2}/4 is normal to the plane 2x + 3y + z at the points with minimum
and maximum distance. |

9) | Suppose that corners of the box have coordinates (±x, ±y, ±z), where x, y and z are positive.
These corners will lie on the ellipse (otherwise, the box could be made bigger). Thus, you are
asked to maximize the volume, 8xyz, subject to the constraint x^{2/4} + y^{2/9} + z^{2/4} = 1. The
maximum occurs where x = 2 3^{-1/2}, y = 3^{1/2},
z = 2 3^{-1/2}. |

10) | An open-top rectangular box of side length x, y, and z (height) has volume xyz and surface area
that is equal to xy + 2(xz + yz) = 36. The maximum volume is z=3 ^{1/2}
for x = y = 2 3^{1/2}
and z =3^{1/2}. Meanwhile, an open-top cylindrical box with radius r and height z has volume
r^{2}h and surface area r^{2} + 2rh = 36.
The maximum volume here is 24 (3/)^{1/2} which occurs, when
r = h = (12/)^{1/2}.
Thus, the cylindrical box has 8 () ^{-1/2} times as much volume as the rectangular
one. |

11) | You are being asked to minimize the function 80x + 25y + 15z subject to the constraint that
300x^{2/5}y^{1/2}z^{1/10} = 12,000.
The minimum occurs for x = 10 (768)^{1/10}, y = 40 (768)^{1/10} and z =
40 (786)^{1/10}/3. |

12a) | You are being asked to minimuze the function 35x + 16y subject to
500x^{7/10} y^{1/2} = 40,000. The minimum occurs at x = 32, y = 50. b) You are being asked to maximize 500x ^{7/10}y^{1/2} subject to 35x + 16y = 4800. The maximum
occurs at x = 80, y = 125. |

13a) | At ten months, x = 100, y = 125 so the money being spent is 100x + 120y = 25,000 and the
production is 300x^{1/2} y^{1/3} = 15,000. |

13b) | You are asked to evaluate P at x = 100 and y = 125. The answer is 75. |

13c) | You are asked to evaluate P at x=100 and y=125. The answer is 40. |

13d) | You are asked to maximize 300x^{1/2}y^{1/3} subject to
100x + 120y = 25,000. The maximum
occurs at x = 150, y = 250/3 and equals 7500 2^{5/6} 3^{1/6}. |

13e) | You are asked to evaluate P at x = 150, y = 250/3.
The answer is 25 2^{5/6} 3^{1/6}. |

Last update, 9/20/2000, math21a@fas.harvard.edu |