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Find the min value of $x^2+2y^2$ if x and y are related by the equation $x+2y=1$.

My attempt,

$$x+2y=1$$

$$2y=1-x$$

$$y=\frac{1-x}{2}$$

$$x^2+2y^2=x^2+2(\frac{1-x}{2})^2$$

$$=x^2+2(\frac{1-2x+x^2}{4})$$

$$=\frac{4x^2+2-4x+2x^2}{4}$$

$$=\frac{6x^2-4x+2}{4}$$

Let $y=\frac{6x^2-4x+2}{4}$

$$\frac{dy}{dx}=3x-1$$

When $\frac{dy}{dx}=0$

$3x-1=0$ $$x=\frac{1}{3}$$

When $x=\frac{1}{3},y=\frac{1}{3}$

Therefore, $x^2+2y^2=\frac{1}{3}$

Am I correct?

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  • $\begingroup$ Are x and y natural, integer or just rational? $\endgroup$ – ghosts_in_the_code Mar 19 '15 at 5:45
  • $\begingroup$ @Mathxx: From line #5 to #6, were is your fraction gone? $\endgroup$ – Frieder Mar 19 '15 at 5:58
  • $\begingroup$ @Frieder I've edited the post. Am I correct? $\endgroup$ – Mathxx Mar 19 '15 at 6:07
  • $\begingroup$ @ghosts_in_the_code the question doesn't mention that $\endgroup$ – Mathxx Mar 19 '15 at 6:14
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$$=x^2+2(\frac{1-2x+x^2}{4})$$

$$=4x^2+2-4x+2x^2$$

This step is wrong.

And remember to plug the value of x and y back into the function to find the min value.

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  • $\begingroup$ I've edited the post, am I correct? $\endgroup$ – Mathxx Mar 19 '15 at 6:07
  • $\begingroup$ @Mathxx Yes ${}$ $\endgroup$ – Jack Mar 19 '15 at 6:10
  • $\begingroup$ @Mathxx: Fine! But before you start differentiation, you may cancel by 2. $\endgroup$ – Frieder Mar 19 '15 at 6:12

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