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This is a Max/min problem, I'm trying calculate the shortest distance between the 2 using pythagoras theorem and diffrenciate it in order to calculate the mininmum of the Red line below:

enter image description here

I'm having trouble putting the equation together...am I going the right way about it?

1st I came up with:

$$f(x) - g(x) = (x+10) - (6\sqrt{x})$$

Of course that didn't work, then I tried ($R$ being the distance of the red line)

$$R = (x+10)^2 - (6\sqrt{x})^2$$

Diffrenciated that and I still didn't get the right answer...I had a feeling both are incorrect, a hint would be brilliant, apologies for asking for many of these questions recently! just getting my brain into shape for September, it's been 2 years since I've done this kinda stuff!

Correct answer: $\dfrac{1}{\sqrt{2}}$

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  • $\begingroup$ Did you take Calculus III? I can provide a solution but it requires Calculus III. $\endgroup$ – wythagoras Jun 19 '15 at 17:20
  • $\begingroup$ Are you familiar with vectors and their inner/dot product? $\endgroup$ – gebruiker Jun 19 '15 at 17:22
  • $\begingroup$ Never heard of Calculus III or vectors and their inner/dot product...I just found a pdf file with some difficult max/min problems...maybe they're beyond my level $\endgroup$ – Modrisco Jun 19 '15 at 17:26
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Hint: Since the function $6 \sqrt{x}$ is concave and strictly below the graph of your line, the minimum distance will occur where the derivative of $6 \sqrt{x}$ is equal to the slope of your line. And the distance will be the perpendicular distance from the line.

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The distance from point to line formula says that the distance from $(x_0, y_0)$ to $y - x - 10 = 0$ is $$\left|\dfrac{y_0 - x_0 - 10}{\sqrt{1^2 + 1^2}}\right| $$

In this case, the point is $(x, 6 \sqrt{x})$. Plug that into the formula, then differentiate.

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The distance between the two points $(x,6\sqrt{x})$ and $(y,y+10)$ is equal to $$\sqrt{(x-y)^2+(y+10-6\sqrt{x})^2}$$

Now you can take partial derivatives.

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