已知实数x、y满足x²+5y²-4xy+2x-8y+1=0，求A=3x-2y的最值

Let's work through this problem step by step. We need to find the maximum and minimum values of ( A = 3x - 2y ) where real numbers ( x ) and ( y ) satisfy the constraint ( x^2 + 5y^2 - 4xy + 2x - 8y + 1 = 0 ).

Step 1: Express ( x ) in terms of ( A ) and ( y ) #

First, rearrange the expression for ( A ) to solve for ( x ):
[
A = 3x - 2y \implies x = \frac{A + 2y}{3}
]
This substitution lets us reduce the problem to a single variable (( y )) by plugging ( x ) into the constraint equation.

Step 2: Substitute into the constraint and simplify #

Plug ( x = \frac{A + 2y}{3} ) into the given constraint:
[
\left( \frac{A + 2y}{3} \right)^2 + 5y^2 - 4 \cdot \frac{A + 2y}{3} \cdot y + 2 \cdot \frac{A + 2y}{3} - 8y + 1 = 0
]
To eliminate fractions, multiply every term by 9:
[
(A + 2y)^2 + 45y^2 - 12y(A + 2y) + 6(A + 2y) - 72y + 9 = 0
]
Expand and combine like terms:
[
A^2 + 4Ay + 4y^2 + 45y^2 - 12Ay - 24y^2 + 6A + 12y - 72y + 9 = 0
]
[
25y^2 + (-8A - 60)y + (A + 3)^2 = 0
]
Now we have a quadratic equation in ( y ): ( ay^2 + by + c = 0 ), where:

• ( a = 25 )
• ( b = -(8A + 60) )
• ( c = (A + 3)^2 )

Step 3: Use the discriminant condition for real solutions #

Since ( y ) must be a real number, the discriminant of this quadratic equation must be non-negative (( \Delta \geq 0 )). The discriminant formula is ( \Delta = b^2 - 4ac ):
[
(8A + 60)^2 - 4 \times 25 \times (A + 3)^2 \geq 0
]
Expand and simplify this inequality:
[
64A^2 + 960A + 3600 - 100(A^2 + 6A + 9) \geq 0
]
[
64A^2 + 960A + 3600 - 100A^2 - 600A - 900 \geq 0
]
[
-36A^2 + 360A + 2700 \geq 0
]
Divide both sides by -36 (remember to reverse the inequality sign):
[
A^2 - 10A - 75 \leq 0
]
Factor the quadratic:
[
(A - 15)(A + 5) \leq 0
]
The solution to this inequality is ( -5 \leq A \leq 15 ).

Step 4: Verify the extreme values #

To confirm these values are valid, we can calculate the corresponding ( x ) and ( y ) and check they satisfy the original constraint:

• When ( A = -5 ): ( y = \frac{2}{5} ), ( x = -\frac{7}{5} ). Substituting back into the constraint gives ( 0 = 0 ), so it holds.
• When ( A = 15 ): ( y = \frac{18}{5} ), ( x = \frac{37}{5} ). This also satisfies the original constraint.

Final Result #

• The minimum value of ( A = 3x - 2y ) is -5
• The maximum value of ( A = 3x - 2y ) is 15

内容的提问来源于stack exchange，提问作者Mary