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Question:

Given $x + y = 12$, find the minimum value of $\sqrt{x^2+4} + \sqrt{y^2+9}$?

Key:

I use $y = 12 - x$ and substitute into the equation, and derivative it.

which I got this

$$\frac{x}{\sqrt{x^2+4}} + \frac{x-12}{\sqrt{x^2-24x+153}} = f'(x).$$

However, after that. I don't know how to do next in order to find the minimum value. Please help!

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  • $\begingroup$ Set $f'(x)=0$ and find solutions to a quadratic equation in $x$. Then analyze to see whether one of the local extrema is the absolute minimum. $\endgroup$ – Mark Viola Jun 1 '15 at 4:12
  • $\begingroup$ Thanks, Dr. MV. Could you give me some hint to do simplification? I have a difficulty to do simplification. $\endgroup$ – Evan C Jun 1 '15 at 4:15
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    $\begingroup$ You can actually solve this problem using basic geometry. Let $A = (0,0)$, $B = (x,2)$, $C = (x+y,2+3) = (12,5)$. Then by the triangle inequality $AB+BC \ge AC$, i.e. $\sqrt{x^2+4}+\sqrt{y^2+9} \ge 13$. Equality occurs iff $B$ is on the line segment $AC$, i.e. $x = 24/5$. $\endgroup$ – JimmyK4542 Jun 1 '15 at 4:23
  • $\begingroup$ @EvanC Per your request, I wrote an answer that outlines a way forward. Please let me know how I can improve the answer. I just want to give you the best answer I can. $\endgroup$ – Mark Viola Jun 1 '15 at 4:27
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Setting $f'(x)=0$ gives

$$\frac{x}{\sqrt{x+4}}+\frac{x-12}{\sqrt{x^2-24x+153}}=0 $$

or

$$\frac{x}{\sqrt{x+4}}=-\frac{x-12}{\sqrt{x^2-24x+153}}\tag1$$

Squaring both sides of $(1)$ gives

$$\left(\frac{x}{\sqrt{x^2+4}}\right)^2=\left(\frac{x-12}{\sqrt{x^2-24x+153}}\right)^2$$

which can be rewritten as

$$\frac{x^2}{x^2+4}=\frac{(x-12)^2}{x^2-24x+153} \tag2$$

Thus, writing the numerator of the left-hand side as $x^2=x^2+4-4$ and writing the numerator of the right-hand side as $(x-12)^2=(x-12)^2+9-9$ simplifies to $(2)$

$$\frac{4}{x^2+4}=\frac{9}{x^2-24x+153} \tag3$$

Can you finish from here?

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  • $\begingroup$ It's very clear for me, thanks a lot Dr. MV $\endgroup$ – Evan C Jun 1 '15 at 5:59
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Just another way: $$\sqrt{x^2+4}+\sqrt{y^2+9} \ge \sqrt{(x+y)^2+(2+3)^2} = 13$$ where we have used the triangle inequality.

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  • $\begingroup$ Woah... I'm speechless $\endgroup$ – Jimmy360 Jun 1 '15 at 4:24
  • $\begingroup$ For completeness, equality is indeed attainable at $x = 24/5$. $\endgroup$ – JimmyK4542 Jun 1 '15 at 4:25
  • $\begingroup$ @JimmyK4542 Very necessary. Equality is attainable whenever the origin is on the line between $(x, \pm 2), (-y, \mp 3)$ $\endgroup$ – Macavity Jun 1 '15 at 4:35
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$$\begin{align} f'(x)=0 &\iff \frac{x}{\sqrt{x^2+4}} = \frac{12-x}{\sqrt{x^2-24x+153}}\\ &\implies \frac{x^2}{{x^2+4}} = \frac{x^2 - 24x + 144}{{x^2-24x+153}}\\ &\iff 1 - \frac{4}{{x^2+4}} = 1 - \frac{9}{{x^2-24x+153}}\\ &\iff 4(x^2 -24x + 153) = 9(x^2+4) \end{align}$$

from which you find the roots (if I'm not mistaken, only one of them is valid because of the squaring in the second line).

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You probably know about setting the derivative equal to $0$. The complexity of the equation we get may be discouraging. Your derivative is $$\frac{x}{\sqrt{x^2+4}}-\frac{12-x}{\sqrt{x^2-24x+153}}.$$ I would rather write it as $$\frac{x}{\sqrt{x^2+4}}-\frac{y}{\sqrt{y^2+9}}.$$ Nicer! Set this equal to $0$. So we get $$\frac{x}{\sqrt{x^2+4}}=\frac{y}{\sqrt{y^2+9}}.$$ Square both sides and simplify a bit (cross-multiply). We get $$x^2(y^2+9)=y^2(x^2+4).$$ There is some nice cancellation, and we get $9x^2=4y^2$. The rest should not be difficult.

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  • $\begingroup$ After squaring both sides, you could take their reciprocals and work with the simpler denominators. $\endgroup$ – cxseven Jun 1 '15 at 5:16
  • $\begingroup$ Yes, then we are cancelling $1$'s instead. I was focusing on using "symmetry." Quite often in calculus problems and elsewhere, it helps to keep things symmetrical, even at the cost of extra variables. The Lagrange multipliers method also would maintain symmetry. $\endgroup$ – André Nicolas Jun 1 '15 at 5:26
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the constraint is $x + y = 12.$ at a local extremum of $\sqrt{x^2 + 4} + \sqrt{y^2 + 9},$ the critical numbers satisfy $$dx + dy =0,\quad \frac{x\, dx}{\sqrt{x^2 + 4}} + \frac{y\, dy}{\sqrt{y^2 +9}} = 0 \to x\sqrt{y^2 + 9}=y\sqrt{x^2 + 4} $$ squaring the last equation we have $$9x^2 =4y^2 \to y = \pm\frac32x, x + y = 12 \implies x=24/5, y = 36/5,\\ \sqrt{x^2 + 4} + \sqrt{y^2+9} = 13$$ therefore the global minimum of $$\sqrt{x^2 + 4} + \sqrt{y^2+9} \text{ is } 13.$$

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If you are aware of the rarely seen in action "Minkowski" inequality..then here you go: $\sqrt{x^2+2^2}+\sqrt{y^2+3^2} \geq \sqrt{(x+y)^2+(2+3)^2}=\sqrt{12^2+5^2} = \sqrt{169}=13$, and rest assured that this extrema is achievable.

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