The answer is 516.

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Generalization : If x/a + y/b = 1 then minimum value of (a + b) + √(a² + b²) is 2(x + y + root(2xy))

Dev Sharma
- 5 years, 5 months ago

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I'm just wondering, is there a geometry solution to this problem? My first thought was that $x +y + \sqrt{x^2+y^2}$ could represent the perimeter of a right angled triangle...

Eamon Gupta
- 5 years, 5 months ago

Agreed, but your generalisation is just a rescaling of the original result.

The value of $g(\alpha)$ in my above post is $2\big[\alpha + 1 + \sqrt{2\alpha}\big]$ .

If $\dfrac{a}{x} + \dfrac{b}{y} = 1$ then $\dfrac{1}{\xi} + \dfrac{a^{-1}b}{\eta} = 1$ where $\xi = a^{-1}x$ and $\eta = a^{-1}y$ . Since $x + y + \sqrt{x^2 + y^2} \; = \; a\big(\xi + \eta + \sqrt{\xi^2 + \eta^2}\big)$ we see that the minimum value of $x + y + \sqrt{x^2 + y^2}$ , subject to $\dfrac{a}{x} + \dfrac{b}{y} = 1$ , is just $a$ times the minimum value of $\xi + \eta+ \sqrt{\xi^2 + \eta^2}$ subject to $\dfrac{1}{\xi} + \dfrac{a^{-1}b}{\eta} = 1$ , which is therefore $a \times 2\big[ 1 + a^{-1}b + \sqrt{2 a^{-1}b}\big] \; = \; 2\big[a + b + \sqrt{2ab}\big] \;.$

Mark Hennings
- 5 years, 5 months ago

How do we get to that?

Saurabh Chaturvedi
- 5 years, 5 months ago

Why can't we take x=36, y=252 to get the minimum value.

Anshuman Singh Bais
- 5 years, 5 months ago

Won't an easier solution be using Lagrange Multipliers?

Aditya Agarwal
- 5 years, 5 months ago

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It's do-able, but I am not sure (after having a brief go) if LM will get the answer any more easily. If you can do it, please show me.

Mark Hennings
- 5 years, 5 months ago

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Okay, I will first do it, and surely put it up if it works!

Aditya Agarwal
- 5 years, 5 months ago

Yeah, I tried it and it IS working. I will post it tomorrow.

Aditya Agarwal
- 5 years, 5 months ago

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First I put the restricting condition into the expression to minimize to get one equation with only one variable x. Then I found the minimum graphically and following numerically with XPLORE to value 536.271887242357. As last step I now had to determine which root ends with the same series of digits and I found $\sqrt{1960}$ .

Using these results it was very simple to obtain the suitable shape after separating the 2 roots included in 1960.

Remark: If someone die-hard mathematician does not like my algorithm I can not change! As retired physicist I prefer such ways especially they possess a distinct scientific character!

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The minimum value is actually $492-14\sqrt{10}$

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No it is not. Your value comes from using the negative root $\alpha_-$ , which is not permitted.

Mark Hennings
- 5 years, 5 months ago

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Without redoing this problem to check this, I'll go along with your explanation.

Michael Mendrin
- 5 years, 5 months ago

how is that?

Dev Sharma
- 5 years, 5 months ago

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We must have $y > a=245$ . Putting $y =a + z$ gives $x \,=\, \dfrac{z+a}{z}$ , and so we need to minimize $g(z) \; = \; \dfrac{z+a}{z} + z+a + \sqrt{\big(\dfrac{z+a}{z}\big)^2 + (z+a)^2} \; =\; \big(1 + \dfrac{a}{z}\big)\big(1 + z + \sqrt{z^2 +1}\big)$ for $z > 0$ . Now $g'(z) \; =\; \frac{(z^2-a)\sqrt{z^2+1} + (z^3-a)}{z^2\sqrt{z^2+1}} \;,$ so we need to solve the equation $(z^2-a)\sqrt{z^2 + 1} + (z^3-a) \,=\, 0$ . Solutions of this equation certainly satisfy (by squaring) $\begin{array}{rcl} (z^2 - a)^2(z^2+1) & = & (z^3 - a)^2 \\ (2a-1)z^4 -2az^3 - (a^2 - 2a)z^2 & = & 0 \\ (2a-1)z^2 - 2az - (a^2 - 2a) & = & 0 \\ z & = & \alpha_\pm \; = \; \dfrac{a}{2a-1} \pm \dfrac{a-1}{2a-1}\sqrt{2a} \end{array}$ Since we must have $z > 0$ we deduce that $z \; = \; \alpha_+ \; =\; \dfrac{a}{2a-1} + \dfrac{a-1}{2a-1}\sqrt{2a}$ Indeed, $\alpha_+$ is not only the positive option out of $\alpha_+$ and $\alpha_-$ ; it is the only solution of $g'(z) = 0$ . The other value $\alpha_-$ is the solution of the equation $(z^2-a)\sqrt{z^2+1} - (z^3-a) \,=\, 0$ , and is therefore a solution of the rogue equation that was introduced when we squared the equation $g'(z) = 0$ to remove the square roots. It is easy to see that $g'(z) < 0$ for $0 < z < \alpha_+$ , with $g'(z) > 0$ for $z > \alpha_+$ , and so $z=\alpha_+$ is indeed a minimum for $g(z)$ .

With $a = 245$ this gives $z = \dfrac{7}{489}\big(35 + 244\sqrt{10}\big)$ and $g(z) = 492 + 14\sqrt{10}$ .