# Minimum value of a function

I want to know how to calculate the minimum value of the following function and similar functions:

$y = \sqrt{(x^2+8x+25)}+ \sqrt{(x^2-8x+25)}$

I know I can just plot the graph in a calculator or computer, but is there any way to calculate it manually? Can somebody help me with this/suggest some resource?

Note by Labib Rashid
5 years, 11 months ago

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Think this as the sum of distance from the point (x,0) to (4,3) and (-4,-3). To minimize, (x,0) has to be on the line passing through (4,3) and (-4,-3).

- 5 years, 11 months ago

Using $2$ Right angle $\triangle$, Like $\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\geq \sqrt{(a+b)^2+(c+d)^2}$

Where $a,b$ denote base and height of one $\triangle$ and $c,d$ denote base and height of other $\triangle$

and these $2-\triangle$ are in opposite directions. and equality hold when $\displaystyle \frac{a}{b} = \frac{c}{d}$

So $\sqrt{(x+4)^2+3^2}+\sqrt{(4-x)^2+3^2}\geq \sqrt{(x+4+4-x)^2+(3+3)^2}= 10$

and equality hold when $\displaystyle \frac{x+4}{3} = \frac{4-x}{3}\Rightarrow x = 0$

- 5 years, 11 months ago

Thanks. Could you suggest where I can find some resources on the idea you used here?

- 5 years, 11 months ago

As far as I can tell, he used Minkowski's inequality of square root sums. But I think George G.'s method is better.

- 5 years, 10 months ago

Generally, differentiate and set equal to zero. For this one I got a minimum at x=0 and y=10. So I conjecture that any equation in this form (just swap out both the eights or both the 25's for other numbers) will have a minimum at x=0.

- 5 years, 11 months ago

We can write,

$y=\sqrt{(x+4)^2+(0+3)^2} + \sqrt {(x-4)^2+(0-3)^2}$

Let $A=(x,0) ; B=(-4,-3) ; C=(4,3)$

So, $y=AB+AC$

Using Triangular Inequality, $AB+BC \geq BC = \sqrt{(4+4)^2+(3+3)^2}=10$

So the minimum value of $y$ is $\fbox{10}$

- 4 years, 9 months ago

just squaring on both times twice we can get a quadratic equation ..this is an alternate way to solve this problem and but it is a tedious one.

- 5 years, 11 months ago

differentiate the function and equate it to zero. The value of 'x' is the local minima.

- 5 years, 11 months ago