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If $x$ and $y$ real numbers such that $4x^2 +y^2 = 4x - 2y +7$ what is the max value of $5x + 6y$?

Taken from the 2017 IMC

Im not sure how to approach this problem

Tyrone
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6 Answers6

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Demand the values of $z$ such that the line $$5x+6y=c~~~~(1)$$ touches the conic $$4x^2+y^2-4x+2y-7=0,~~~~(2)$$ put $y=(c-5x)/6$ in (2) , we get $$\frac{169}{36}x^2-(17/3+5c/18)x+c^2/36-7==0.~~~~(3)$$ Let us demand that $$B^2=4AC.$$ we get $$c^2+7z-368=(z-16)(c+23)=0.$$ These two values $c=16,-23$ are two values of $c$ when the line (1) touches the ellipse (2). These are the required maximum (16) and minimum (-23) values of $5x+6y$ under the condition (2).

Note: Given the condition as a conic, $f(x,y)=0$, the line $ax+by=c$ will have the intercept $c$ max/min when it touches the conic.

Z Ahmed
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    Equation (1) describes a plane, not a line. – amd Jul 05 '19 at 23:53
  • @amd (1) is a line and (2) is a curve and it is 2D. Not 3D, why invoke 3D?! – Z Ahmed Jul 06 '19 at 07:01
  • If you meant for $z$ to be an arbitrary constant, convention would have you use a different letter: $z$ strongly suggests that it is a coordinate in $\mathbb R^3$. That aside, the task is to find a constrained maximum of $5x+6y$, which is naturally equivalent to finding the point on the intersection of the plane $z=5x+6y$ with the cylinder $4x^2+y^2=4x-2y+7$ that has the greatest $z$-coordinate. – amd Jul 06 '19 at 07:20
  • That aside, this is a nice solution. – amd Jul 06 '19 at 07:29
  • @amd Thanks, will I get an upvote? – Z Ahmed Jul 06 '19 at 09:20
  • Yes you will! +1 – nonuser Jul 06 '19 at 15:11
  • I’d have no hesitation to upvote if you explained why the tangent lines produce the extrema. – amd Jul 06 '19 at 17:30
  • @amd Tangent is critical to a curve so is max/min of a linear function subject tp the quadraric condition. It is like we put $y=mx+c$ in a quaratic curve and demand B^2-4AC (coincident roots: tangency) tp get correct value or expression of c. – Z Ahmed Jul 07 '19 at 06:56
  • In the answer, not in a comment to it. – amd Jul 07 '19 at 07:00
  • @amd I have now added an appropriate note in the end of my solution itself, thanks. – Z Ahmed Jul 12 '19 at 02:51
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We can rewriting the equation, we get $$4\left(x-\frac{1}{2}\right)^2 + (y+1)^2 = 9$$ Thus, our curve is an ellipse. We can parameterize this curve as $x(t) = \frac{1}{2} + \frac{3}{2}\cos t$ and $y(t) = -1 + 3\sin t$. Plugging these into the quantity to be maximized, we obtain $-\frac{7}{2}+\frac{15}{2}\cos t + 18 \sin t$. This implies that we have a local maximum at $\tan t = \frac{12}{5}$. It follows $\cos t = \frac{5}{13}$ and $\sin t = \frac{12}{13}$ and our maximum is $16$.

You can verify the result here.

Sohom Paul
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Note that\begin{align}4x^2 +y^2 = 4x - 2y +7&\iff(2 x-1)^2+(y+1)^2-9=0\\&\iff\left(\frac{2x-1}3\right)^2+\left(\frac{y+1}3\right)^2=1,\end{align}So, the pairs $(x,y)$ such that $4x^2 +y^2 = 4x - 2y +7$ are all of the form $\left(\frac{1+3\cos\theta}2,\frac{-1+3\sin\theta}2\right)$. And, for $\theta\in\mathbb R$,\begin{align}5\frac{1+3\cos\theta}2+6\frac{-1+3\sin\theta}2&=18 \sin (\theta )+\frac{15 \cos (\theta )}{2}-\frac{7}{2}\\&=\frac12\left(36\sin(\theta)+15\cos(\theta)-7\right)\\&=\frac{39}2\left(\frac{36}{39}\sin(\theta)+\frac{15}{39}\cos(\theta)\right)-\frac72.\end{align}Since $\left(\frac{36}{39}\right)^2+\left(\frac{15}{39}\right)^2=1$, there is some $\alpha\in\mathbb R$ such that $\cos(\alpha)=\frac{36}{39}$ and that $\sin(\alpha)=\frac{15}{39}$. Therefore$$5\frac{1+3\cos\theta}2+6\frac{-1+3\sin\theta}2=\frac{39}2\cos(\alpha-\theta)-\frac72$$and the maximum of your expression is $\frac{39}2-\frac72=16$.

José Carlos Santos
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Solve $y^2+2y+(4x^2-4x-7)=0\,$ using the quadratic formula to get

$y=-1\pm2\sqrt{2+x-x^2} =-1\pm2\sqrt{(1+x)(2-x)}$

[note: for $y$ to be real we need $-1\le x\le 2].$

We want to maximize $f(x)=5x+6y=5x-6\pm12\sqrt{2+x-x^2},$

so solve $f’(x)=5\pm\dfrac{6(1-2x)}{\sqrt{2+x-x^2}}=0.$

Thus $25(2+x-x^2)=36(1-2x)^2\implies 169x^2-169x-14=(13x+1)(13x-14)=0,$

so $x=-\dfrac 1{13}$ or $\dfrac{14}{13}$.

Trying these values (as well as endpoints $-1$ and $2$ too) yields the maximum of $f(x)$ is $16,$

when $x=\dfrac{14}{13}$ and $y=\dfrac{23}{13}.$

J. W. Tanner
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You could always bash this out using Lagrange multipliers: Set $$L(x,y)=5x+6y-\lambda(4x^2+y^2-4x+2y-7)$$ and compute $$\nabla L = (5-8\lambda x-4\lambda,6-2\lambda y+2\lambda).$$ You can solve $\nabla L=0$ for $x$ and $y$, producing $x={4\lambda-5\over8\lambda}$ and $y=-{\lambda+3\over\lambda}$. Substitute into the conic constraint equation and solve the resulting quadratic equation for $\lambda$, then choose the one that produces the maximum value of $5x+6y$.

Incidentally, the Lagrange multiplier method provides a motivation for this other solution: the constrained extrema of $f(x,y)=5x+6y$ occur where its gradient is normal to the curve, but this is equivalent to finding where level curves of $f$, which are straight lines, are tangent to the curve.

amd
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The curve $4x^2 +y^2 = 4x - 2y +7$ is an ellipse. The objective function $z=5x + 6y$ must be maximized. Consider the contour lines: $$y=-\frac56x+\frac z6$$ The tangent line to the ellipse must have the slope: $$8x_0+2y_0y'=4-2y' \Rightarrow y'=\frac{2-4x_0}{y_0+1}=-\frac56 \Rightarrow y_0=\frac{24x_0-17}{5}$$ Substitute this to equation of ellipse: $$4x^2+\left(\frac{24x-17}{5}\right)^2=4x-2\cdot \frac{24x-17}{5}+7 \Rightarrow \\ x=-\frac{1}{13}, \frac{14}{13}.$$ Hence, max $z(\frac{14}{13},\frac{23}{13})=5\cdot \frac{14}{13}+6\cdot \frac{23}{13}=16$.

Can you find the minimum value?

farruhota
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