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?

• Are x and y natural, integer or just rational? – ghosts_in_the_code Mar 19 '15 at 5:45
• @Mathxx: From line #5 to #6, were is your fraction gone? – Frieder Mar 19 '15 at 5:58
• @Frieder I've edited the post. Am I correct? – Mathxx Mar 19 '15 at 6:07
• @ghosts_in_the_code the question doesn't mention that – Mathxx Mar 19 '15 at 6:14

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

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

This step is wrong.

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

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