1996 AMC 12 第 25 题

Question

已知

x^2+y^2=14x+6y+6

，

3x+4y

的最大可能值是多少？

Given that

x^2+y^2=14x+6y+6,

what is the largest possible value that

3x+4y

can have?

72

73

74

75

76

配方，确定圆心和半径

Complete the square to identify the circle’s center and radius

3x+4y

的最大值等于它在圆心处的值加上半径乘以

\sqrt{3^2+4^2}

The maximum of

3x+4y

is its value at the center plus the radius times

\sqrt{3^2+4^2}

配方得

(x-7)^2+(y-3)^2=64\text{。}

在圆心处，

3x+4y=3(7)+4(3)=33

。移动

8

个单位，方向取为向量

(3,4)

的方向，表达式会增加

8\sqrt{3^2+4^2}=40

。最大值为

73

，所以正确答案是 B。

Completing the square gives

(x-7)^2+(y-3)^2=64.

At the center,

3x+4y=3(7)+4(3)=33.

Moving

8

units in the direction of the vector

(3,4)

increases the expression by

8\sqrt{3^2+4^2}=40.

The maximum is

73,

so the correct answer is B.

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