Help: Minimum

Find the value of x such that the value of x²6x+25\sqrt{x²-6x+25} + x²+10x+169\sqrt{x²+10x+169} is at its minimum.

Note by Vincent Lopez
2 months, 3 weeks ago

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x=1x=1

James Watson - 2 months, 2 weeks ago

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Mind sharing your solution bro? I'm kinda stuck with this

x²6x+9+16 \sqrt{x²-6x+9+16} + x²+10x+25+144 \sqrt {x²+10x+25+144}

=(x3)²+4² \sqrt{(x-3)²+4²} + (x+5)²+12² \sqrt{(x+5)²+12²}

I know solving the x-intercept of the line passing through the points (3,4) (3,-4) and (5,12) (-5,12) or (3,4) (3,4) and (5,12) (-5,-12) will give the answer but I don't know why is it like that.

Vincent Lopez - 2 months, 2 weeks ago

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i used calculus for this; idk the algebraic way. If you want to know the calculus route i can share it

James Watson - 2 months, 2 weeks ago

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@James Watson Yes bro, if you don't mind.

Vincent Lopez - 2 months, 2 weeks ago

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@Vincent Lopez You can find the derivative of the function and set that equal to 0. Then solve for xx. To make sure that it is a minimum point, take the second derivative of the function to see whether it is concave up on the stationary point (concave up means that d2ydx2>0\displaystyle \frac{d^2 y}{dx^2} > 0 )

James Watson - 2 months, 2 weeks ago

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@James Watson f(x)=x²6x+25+x²+10x+169f(x)=\sqrt{x²-6x+25} +\sqrt{x²+10x+169}

f(x)=x3x²6x+25+x+5x²+10x+169=0f'(x)=\frac{x-3}{\sqrt{x²-6x+25}} +\frac{x+5}{\sqrt{x²+10x+169}}=0

x+5x²+10x+169=x3x²6x+25\frac{x+5}{\sqrt{x²+10x+169}}=-\frac{x-3}{\sqrt{x²-6x+25}}

Squaring both sides of the equation

(x+5)²x²+10x+169=(3x)²x²6x+25\frac{(x+5)²}{x²+10x+169}=\frac{(3-x)²}{x²-6x+25}

The reciprocals of equals are equal

x²+10x+169(x+5)²=x²6x+25(3x)²\frac{x²+10x+169}{(x+5)²}=\frac{x²-6x+25}{(3-x)²}

(x+5)²+12²(x+5)²=(x3)²+4²(3x)²\frac{(x+5)²+12²}{(x+5)²}=\frac{(x-3)²+4²}{(3-x)²}

1+12²(x+5)²=1+4²(3x)²1+\frac{12²}{(x+5)²}=1+\frac{4²}{(3-x)²}

12²(x+5)²=4²(3x)²\frac{12²}{(x+5)²}=\frac{4²}{(3-x)²}

Get the square root of both sides

12x+5=43x\frac{12}{x+5}=\frac{4}{3-x}

x=1\boxed{x=1}

Is this the right way bro?

Vincent Lopez - 2 months, 2 weeks ago

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@Vincent Lopez yeah. Don't forget to check if it is a minimum point with the second derivative!

James Watson - 2 months, 2 weeks ago

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@James Watson When I thought I was done but completely forgotten the most important part, my eyes can't keep up anymore. I'll check it tomorrow bro, thanks for helping me again :)

Vincent Lopez - 2 months, 2 weeks ago

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@Vincent Lopez haha no problem!

James Watson - 2 months, 2 weeks ago

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@James Watson f(x)=x²6x+25+x²+10x+169f(x)=\sqrt{x²-6x+25} +\sqrt{x²+10x+169}

f(x)=x3x²6x+25+x+5x²+10x+169f'(x)=\frac{x-3}{\sqrt{x²-6x+25}} +\frac{x+5}{\sqrt{x²+10x+169}}

f(x)=16x²6x+25+144x²+10x+169f''(x)=\frac{16}{\sqrt{x²-6x+25}} +\frac{144}{\sqrt{x²+10x+169}}

I don't quite get it bro, do I need find out if f(1)>0f''(1)>0?

Vincent Lopez - 2 months, 2 weeks ago

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@Vincent Lopez yes

James Watson - 2 months, 2 weeks ago

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@James Watson Thanks bro for helping me all throughout the solution :) I don't even know how to use emoji. hahaha

Vincent Lopez - 2 months, 2 weeks ago

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@Vincent Lopez lol no problem mate

James Watson - 2 months, 2 weeks ago

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I think you are almost done. Let A(3,4),B(5,12),P(x,0)A(3,-4), B(-5,12), P(x,0), then AP+BPABAP+BP \geq AB by triangle inequality, and the equality holds when PP lies on segment ABAB.

X X - 2 months, 2 weeks ago

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@X X Why is it that point A has an ordinate of 4-4 instead of +4+4?

Vincent Lopez - 2 months, 2 weeks ago

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@Vincent Lopez If the y coordinate is 4, the distance from P to A won't change. However, P cannot be on segment AB, so the triangle inequality's equality won't hold.(i.e. Choose A and B such that A and B are on different side of the X axis

X X - 2 months, 2 weeks ago

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@X X Took me long to understand, thanks man. :)

Vincent Lopez - 2 months, 2 weeks ago

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?-\infty? @Vincent Lopez

Percy Jackson - 2 months, 3 weeks ago

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@Percy Jackson it's 1 but I don't know how to solve it.

P.S edited already, should be help.

Vincent Lopez - 2 months, 2 weeks ago

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