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Show that the sum of the $x$-, $y$- and $z$-intercepts of any tangent plane to the surface $\sqrt x +\sqrt y +\sqrt z = 1$ is a constant.

I tried making an equation: $z = (1-\sqrt{x}+\sqrt{y})^2$ and I followed the standards steps of finding an tangent plane through the partial derivatives with respect to $x$ and $y$. I let $x_0, y_0, z_0$ be any point on the surface. To find each intercept, I substitute $0$ for the other two variables. All I got are complicated equations.

Any ideas?

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You need to write equation of plane in implicit form ie $F(x,y,z) = \sqrt x + \sqrt y + \sqrt z - 1 = 0$. Then you can write equation of tangent as $F_x(x,y,z) (x-x_0) + F_y (x,y,z)(y - y_0) + F_z (x,y,z)(z-z_0) = 0 $. And finally substitute suitable zeros (eg y =0, z=0 for x-intercept) to find all 3 intercepts. When you add them it would come as some function of $(\sqrt x + \sqrt y + \sqrt z)$, hence constant.

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I got: $\sqrt{x0}(\sqrt{y0}+\sqrt{z0}+1)+\sqrt{y0}(\sqrt{x0}+\sqrt{z0}+1)+ \sqrt {z0}(\sqrt{y0}+\sqrt{x0}+1)$ What am i doing wrong? – uohzxela Oct 28 '12 at 9:04

I wanted to post a question on LieX's answer, but couldn't find a way to do so. Anyway, i have additional query with regards to LieX's solution.

$$F(x,y,z)=√x+√y+√z−1=0;$$ $$f_x = 1/2√x$$ $$f_y = 1/2√y$$ $$f_z = 1/2√z$$

Hence, the tangent plane equation = $$(1/2√x)(x-x_0) + (1/2√y)(y-y_0) + (1/2√z)(z-z_0)$$

How should I continue from here? M I on the right track??

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Remember to plug the coordinates of the point $(x_0, y_0, z_0)$ into the equations for the partials. With this correction, the tangent plane equation is the expression that you wrote down set equal to zero. Set $x = 0, y = 0$ (to get the z-intercept) and use the tangent plane equation together with Then use the fact that $\sqrt{x_0} + \sqrt{y_0} + \sqrt{z_0} = 1$ to solve for $z$. Similarly, solve for the $x$ and $y$ intercepts. There may be a much more elegant approach, but this one should work... – Jonah Sinick Oct 29 '12 at 4:49

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