Calculus - Find The Minimum Value Of $y = \sqrt {{x^2} + 4x + 13} + ...

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Learn more about Teams Find the minimum value of $y = \sqrt {{x^2} + 4x + 13} + \sqrt {{x^2} - 8x + 41} $ Ask Question Asked 3 years ago Modified 3 years ago Viewed 730 times 2 $\begingroup$

Find the minimum value of $y = \sqrt {{x^2} + 4x + 13} + \sqrt {{x^2} - 8x + 41} $ .

My approach is as follow

$y = \sqrt {{x^2} + 4x + 4 + 9} + \sqrt {{x^2} - 8x + 16 + 25} \Rightarrow y = \sqrt {{{\left( {x + 2} \right)}^2} + 9} + \sqrt {{{\left( {x - 4} \right)}^2} + 25} $

Within the square roots the minimum values are 3 and 5 hence the minimum value should be greeter than 8 but not able to find the actual minimum value.

• calculus
• algebra-precalculus

Share Cite Follow edited Nov 12, 2021 at 14:09 RAHUL 1,56366 silver badges1919 bronze badges asked Nov 12, 2021 at 13:40 Samar Imam ZaidiSamar Imam Zaidi 8,99444 gold badges2727 silver badges7272 bronze badges $\endgroup$ 1

• 4 $\begingroup$ Take points $(-2, 3), (x, 0), (4, -5)$. Value of $x$ that makes these points collinear should be the min $\endgroup$ – Math Lover Commented Nov 12, 2021 at 13:48

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2 Answers 2

Sorted by: Reset to default Highest score (default) Date modified (newest first) Date created (oldest first) 11 $\begingroup$

I'll approach as geometry.

Let three point as $A(-2,-3), P(x,0), B(4,5)$. Then $y=\overline{AP}+\overline{PB}$.

By triangle inequality, $y\ge\overline{AB}=10$.

So the minimum value is $10$.

Share Cite Follow answered Nov 12, 2021 at 13:48 MH.LeeMH.Lee 5,6531010 silver badges3232 bronze badges $\endgroup$ 2

• $\begingroup$ Really nice. (+1) $\endgroup$ – trancelocation Commented Nov 12, 2021 at 15:17
• $\begingroup$ Make that +2. You beat me to the punch! $\endgroup$ – Oscar Lanzi Commented Nov 12, 2021 at 16:30

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I was going to do a geometrical solution, but looks like that's already been done.

As for a calculus solution, we have $$\frac{dy}{dx}=\frac{x+2}{\sqrt{x^2+4x+13}}+\frac{x-4}{\sqrt{x^2-8x+41}}$$ As, you have already shown $\sqrt{x^2+4x+13}>0$ and $\sqrt{x^2-8x+41}>0$ from completing the square. Hence, the only critical points occur when the RHS is equal to zero. $$\frac{x+2}{\sqrt{x^2+4x+13}}+\frac{x-4}{\sqrt{x^2-8x+41}}=0$$ $$\frac{x+2}{\sqrt{x^2+4x+13}}=-\frac{x-4}{\sqrt{x^2-8x+41}}$$ $$\frac{x^2+4x+4}{x^2+4x+13}=\frac{x^2-8x+16}{x^2-8x+41}$$ $$\frac{x^2+4x+4}{x^2+4x+13}-1=\frac{x^2-8x+16}{x^2-8x+41}-1$$ $$\frac{-9}{x^2+4x+13}=\frac{-25}{x^2-8x+41}$$ $$9x^2-72x+369=25x^2+100x+325$$ $$16x^2+172x-44=0$$ $$4x^2+43x-11=0$$ $$(4x-1)(x+11)=0$$ $$x=-11,\frac{1}{4}$$ However, since we squared both sides of the equation, we expect one of the solutions to be extranneous. Note that if $x=-11$, then $x+2$ and $x-4$ are both negative. Hence, $\frac{x+2}{\sqrt{x^2+4x+13}}$ and $\frac{x-4}{\sqrt{x^2-8x+41}}$ will have the same sign. Then of course we could not have $\frac{x+2}{\sqrt{x^2+4x+13}}=-\frac{x-4}{\sqrt{x^2-8x+41}}$. So our solution is $x=\frac{1}{4}$. It is not hard to deduce that this is a minima.

So the minimum value of $y$ is $$\sqrt{\frac{1}{16}+14}+\sqrt{\frac{1}{16}+39}$$ $$=\sqrt{\frac{225}{16}}+\sqrt{\frac{625}{16}}$$ $$=\frac{15}{4}+\frac{25}{4}$$ $$=\boxed{10}$$

Share Cite Follow answered Nov 12, 2021 at 14:08 Alan AbrahamAlan Abraham 5,19277 silver badges2121 bronze badges $\endgroup$ Add a comment |

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