Best solutions to - "Given that 2x2−4xy+6y2=9, find the biggest and lowest..."

Landen Barber

Answered question

2022-03-27

Given that

2 x^{2} - 4 x y + 6 y^{2} = 9

, find the biggest and lowest value of

2 x - y

Jermaine Lam

Beginner2022-03-28Added 11 answers

Step 1

2 x^{2} - 4 x y + 6 y^{2} = 9

and let

k = 2 x - y \Rightarrow y = 2 x - k

2 x^{2} - 4 x (2 x - k) + 6 {(2 x - k)}^{2} = 9

2 x^{2} - 8 x^{2} + 4 x k + 6 (4 x^{2} + k^{2} - 4 x k) = 9

- 6 x^{2} + 4 x k + 24 x^{2} + 6 k^{2} - 24 x k - 9 = 0

18 x^{2} - 20 k x + 6 k^{2} - 9 = 0

For real roots, Discriminant

\geq 0

400 k^{2} - 4 \cdot 18 (6 k^{2} - 9) \geq 0

100 k^{2} - 108 k^{2} + 162 \geq 0

- 8 k^{2} + 162 \geq 0 \Rightarrow k^{2} - \frac{81}{4} \leq 0

- \frac{9}{2} \leq k \leq \frac{9}{2}

Kamora Campbell

Beginner2022-03-29Added 13 answers

Step 1
The trick is to notice that:

4 {(2 x - y)}^{2} = 9 (2 x^{2} - 4 x y + 6 y^{2}) - 2 {(x - 5 y)}^{2} \leq 9 (2 x^{2} - 4 x y + 6 y^{2}) = 81

It follows that

- \frac{9}{2} \leq 2 x - y \leq \frac{9}{2}

The minimum is attained when

(x, y) = (- d \frac{5}{2}, - d \frac{1}{2})

and the maximum when

(x, y) = (d \frac{5}{2}, d \frac{1}{2})

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