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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?

Thanks in advance.

Note by Labib Rashid
3 years, 10 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).

George G - 3 years, 10 months ago

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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 \(\displaystyle \frac{x+4}{3} = \frac{4-x}{3}\Rightarrow x = 0\)

Jagdish Singh - 3 years, 10 months ago

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

Labib Rashid - 3 years, 10 months ago

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

Mursalin Habib - 3 years, 9 months ago

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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.

Bob Krueger - 3 years, 10 months ago

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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 \(\fbox{10}\)

Ahmed Arup Shihab - 2 years, 8 months ago

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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.

Pradaab Cp - 3 years, 10 months ago

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

Arjun Shankar - 3 years, 10 months ago

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