Minimum value of a function

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

y=(x2+8x+25)+(x28x+25)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
6 years, 1 month ago

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3 votes

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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 - 6 years, 1 month ago

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Using 22 Right angle \triangle, Like a2+b2+c2+d2(a+b)2+(c+d)2\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\geq \sqrt{(a+b)^2+(c+d)^2}

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

and these 22-\triangle are in opposite directions. and equality hold when ab=cd\displaystyle \frac{a}{b} = \frac{c}{d}

So (x+4)2+32+(4x)2+32(x+4+4x)2+(3+3)2=10\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 x+43=4x3x=0\displaystyle \frac{x+4}{3} = \frac{4-x}{3}\Rightarrow x = 0

jagdish singh - 6 years, 1 month ago

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

Labib Rashid - 6 years, 1 month 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 - 6 years 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 - 6 years, 1 month ago

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We can write,

y=(x+4)2+(0+3)2+(x4)2+(03)2y=\sqrt{(x+4)^2+(0+3)^2} + \sqrt {(x-4)^2+(0-3)^2}

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

So, y=AB+ACy=AB+AC

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

So the minimum value of yy is 10\fbox{10}

Ahmed Arup Shihab - 4 years, 11 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 - 6 years, 1 month ago

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

Arjun Shankar - 6 years, 1 month ago

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