$4x^2-5xy+4y^2=19$ with $Z=x^2+y^2$, find $\dfrac1{Z_{\max}}+\dfrac1{Z_{\min}}$

Question

I have no idea how to approach this question at all. I've tried to find the maximum and minimum of the quadratic but i am too confused on what to do afterwards.

Answer 1

Let $x=\sqrt{z}\cos B, y=\sqrt{z}\sin B$, hence we get $$4z-\frac{5z}{2}\sin\left (2B \right )=19$$ for $-1\leq \sin\left ( 2B \right )\leq 1$ and $\displaystyle z=\frac{19}{4-\dfrac{5}{2}\sin\left ( 2B \right )}$. Now the answer will follow.

Answer 2

Without Lagrange multipliers (but with trigonometry). Use polar coordinates:

$x=\sqrt Z\cos\theta$

$y=\sqrt Z\sin\theta$

$xy=Z\sin\theta\cos\theta=\frac{Z}{2}\sin(2\theta)$

From your equation you get $\frac{1}{z}=\frac{4}{19}-\frac{5\sin(2\theta)}{38}$

As $\sin$ takes values between $-1$ and $1$:

So $\frac{1}{z_{min}}+\frac{1}{z_{max}}=2\cdot\frac{4}{19}=\frac{8}{19}$

Answer 3

$$4x^2-5xy+4y^2-19=0$$

By quadratic formula we have,

$$x=\frac{1}{8} (5 y\pm \sqrt{304-39y^2})$$

So,

$$z=y^2+\frac{1}{64} (5 y\pm \sqrt{304-39y^2})^2$$

Now take derivative, set it to zero, and examine sign changes.

Answer 4

• From the given equation $\,xy = \cfrac{4(x^2+y^2) - 19}{5} = \cfrac{4 Z - 19}{5}\,$. But for $\forall x,y \in \mathbb{R}$ the inequality holds $\,xy \le \cfrac{1}{2}(x^2+y^2)\,$ so $\,\cfrac{4 Z - 19}{5} \le \cfrac{Z}{2} \iff \boxed{Z \le \cfrac{38}{3}}\,$

• $0 \le (x+y)^2 = x^2+y^2+2xy = Z + 2 \cfrac{4 Z - 19}{5}=\cfrac{13 Z - 38}{5} \implies \boxed{Z \ge \cfrac{38}{13}}$

$Z_{min}=\cfrac{38}{13}$ is attained for $x=-y=\sqrt{\cfrac{19}{13}}$ and $Z_{max}=\cfrac{38}{3}$ is attained for $x=y=\sqrt{\cfrac{19}{3}}\,$.