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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.

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    Which calculus course is this for? Have you learned about the method of Lagrange multipliers? – 2017-01-10
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    I am learning this on my own, we haven't been taught calculus yet at school and no I haven't learned about Lagrange multipliers. Could you expand please. – 2017-01-10
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    They are a technique for optimizing of a function such as $Z=x^2+y^2$, subject to a constraint like $4x^2-5xy+4y^2=19$. The idea is that your constraint forms a curve, and $Z$ is only optimized on that curve if its direction of greatest increase is perpendicular to the curve at that point (since otherwise you could move along the curve to increase or decrease $Z$). My personal favourite introduction to Lagrange multipliers is in Paul's online math notes: http://tutorial.math.lamar.edu/Classes/CalcIII/LagrangeMultipliers.aspx – 2017-01-10

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

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.

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  • 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 $x=y=\sqrt{\cfrac{19}{3}}\,$.

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

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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.

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    How would you follow on from this, sorry i'm not very good at calculus. – 2017-01-10
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    @Julian Lopez using $-1\leq \sin\left ( 2B \right )\leq 1$ and you will get the maximum and minimum of $z$. – 2017-01-10