Extrema with constraints

例 15   A cylindrical can contains $\upsilon \, c. c.$ of juice. Find the radius of its two ends and its height so that it is constructed with the least amount of material.

Solution. Let the radius of either end be x cm. and the height be y cm. respectively. Then the surface area is $2\pi x(x + y)$ and the volume is $\pi x^{2}y$. We want to minimize the surface area subject to the condition that the volume is the constant $\upsilon$. This leads us to consider the critical values of the function

$$\Lambda(x,y,\lambda) = x(x + y) + \lambda(x^{2}y - \frac {\upsilon}{\pi}) .$$

Equating the partial derivatives of $\Lambda$ with respect to x, y and $\lambda$ to 0, we get

$$2x + y + 2\lambda xy = 0,\eqno(1)$$

$$x + \lambda^{2} = 0,\eqno(2)$$

$$x^{2}y-\frac{\upsilon}{\pi}=0, \eqno(3)$$

From (3) we see that $x\neq 0$. Hence (2) implies that $\lambda x=-1$. Substituting into (1), we get 2x+y-2y=0, or y=2x. This, together with (3), gives

$$x =\sqrt[3]{\frac{\upsilon }{2\pi } }, \quad y=\sqrt[3]{\frac{4 \upsilon }{\pi} }$$

例 16   Find the maximum of the product of n non-negative numbers when their arithmetic mean is given.

Solution. Let the n non-negative numbers be $x_{1}, x_{2},\ldots , x_{n}$ and let their arithmetic mean be a. Our task is to maximize the function

$$x_{1} \,x_{2} \, \cdots \, x_{n}$$

under the constrait

$$x_1+x_2+\, \cdots \, +x_n-na=0 .$$

Consider the function

$$\Lambda (x_{1}, x_{2},\ldots,x_{n},\lambda)=x_1x_2 \ldots x_n+\lambda (x_{1} + x_{2} + \ldots + x_{n} -na) .$$

To evaluate its critical values, solve the following equations which are obtained by equating its partial derivatives to 0.

$$x_{1} + x_{2} + \cdots + x_{n} - na = 0,\eqno(4)$$

$$x_{1} x_{2} \cdots x_{i-1} x_{i+1} \cdots x_{n} =-\lambda, i = 1, 2,\ldots , n.\eqno(5)$$

Multiply the n equations in (5) together, and we obtain

$$x_{1}x_{2} \cdots x_{n} = (-\lambda)^{\frac{n}{n-1}}.\eqno(6)$$

As the maximum will not occur where one of the x_i is 0, it is legal to divide (6) with each of the equations in (5), and we obtain $x_{1} = x_{2} = \cdots = x_{n}$. Substituting into the constraint, we get that the common value of the x_i is a. This result may be formulated as the following theorem:

Strictly speaking, as a complete proof of this fact, the above discussion contains a gap, because we do not know whether the critial point yields a maximum, a minimum or a saddle point. Though this gap may be mended by some compactness argument, it lies beyond the scope of the present course. For general educational purpose, we present below an algebraic proof of Theorem 9: Proof (Cauchy). The case n = 2 is done in the following way:

$$0\leq (x_{1}^{\frac{1}{2}}-x_2^{\frac{1}{2}})^{2} = x_{1} - {2}(x_1x_2)^{\frac{1}{2}} + x_{2} ,$$

with wquality valid for x₁=x₂ . Assume that we have the result for n=2^k , and let 2^k+1 non-negative numbers $x_{1},x_{2},\ldots,x_{2^{k+1}}$ be given. Then

習$\quad$題

1.

若 $n\geq1$, $x\geq0$, $y\geq0$, 證明 $$\frac{x^n+y^n}{2}\geq(\frac{x+y}{2})^n$$.
提示. 在 x+y= 常數的條件下求函數 $$z=\frac{1}{2}(x^n+y^n)$$ 的極小值.

2.

Let $p_{i}=(a_{i}, b_{i}) \in \mathbb{R} ^2, i =1, 2,\ldots , n.$Find the point $z \in \mathbb{R} ^2$ where $\sum_{i=1}^{n} \Vert z - p_{i} \Vert^2$ achieves minimum.

3.

在 ellipsoid $$\frac{x^2}{a^2}+\frac{y^2}{b^2 }+\frac{z^2}{c^2}=1$$內嵌入有最大體積的直角平行六面體.

4.

設製造貨櫃底的材料每平方公尺價值 1200 元, 其邊每平方公尺為 700 元. 欲使貨櫃容積為 14 立方公尺, 問貨櫃之長, 寬, 高各如何始可使材料費為最小?

5.

求 (7,-2,3) 到曲面 x²+4y²+z²-20x+16y+16=0 的最短和最長距離.

6.

在間隔 [1, 3] 中用一次函數 ax + b 來近似地代替函數 x²使得絕對偏差

$$\max\{ \vert x^2 -(ax+b)\vert:1 \leq x \leq 3 \}$$

為最小.

7.

(David A. Smith) 設

f(x, y)=e^-y2(2x³-3x²+1)+e^-y(2x³-3x²).

(a) 試證 f(x,y) 的唯一的 critical point 在 (0,0), 且 f(0,0)=1 是它的相對極大值.
(b) 驗證 $f(2,0) \geq f(0,0)$ .
註. 從本題可以看出:就定義在全平面上的兩個變數的函數而言, 唯一的 critical point 為相對極大值時仍未必是絕對極大值.