題目

Problem

PART 1: Fill in the blanks.

Work need not be shown, only answers will be graded.
4 points for each blank and 100 points total.
Each answer need to be clearly labeled on the answer sheet.
Use of any device with computer algebra system during the exam will result in zero points.

The shortest distance from the origin to the paraboloid $z=\dfrac{x^2+2y^2-36}{4}$ is $\underline{\quad(19)\quad}$ .

解答

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把距離平方寫成 $x^2,y^2$ 的函數，並連同邊界一起比較。

展開

令 $u=x^2,v=y^2$ ，則 $u,v\ge0$ ，距離平方為

D^2=u+v+\frac{(u+2v-36)^2}{16}.

內部臨界點須同時滿足

1+ rac{u+2v-36}{8}=0,qquad 1+ rac{u+2v-36}{4}=0,

兩式互相矛盾，所以最小值只能在邊界。當 $v=0$ 時，

D^2=u+ rac{(u-36)^2}{16}

的最小值為 $32$ ；當 $u=0$ 時，

D^2=v+ rac{(2v-36)^2}{16} = rac14(v-16)^2+17,

故在 $v=16$ 時取得較小的 $D^2=17$ 。

故最短距離為 $\boxed{\sqrt{17}}$ 。