How to maximize or minimize $2y^2 + x^2$ subject to the constraint $x+y=1$?

Question

How do I maximize or minimize $2y^2 + x^2$ such that $x + y = 1$?

I'm a newbie and don't know how to start.

Answer 1

HINT: let $$f(x,y)=2y^2+x^2$$ under the condition $$x+y=1$$ with $$y=1-x$$ you will have $$f(x,1-x)=2(1-x)^2+x^2$$ and this reduces the Problem in one variable.

Answer 2

1 - Reduce the complexity of your problem by letting $y=1-x$ and substitute that into your equation.

2 - Take the derivative of the simplified express and set that equal to zero. Solve for the roots to find the turning points

3 - In general, if you can't graph the equation, take the second derivative to analyse the gradient of the equation from point 2. By finding the second derivative and inserting values which correspond to just before and just after your turning points, you can deduce whether the point is a minima or a maxima.

Answer 3

By C-S $$x^2+2y^2=\frac{(2+1)(x^2+2y^2)}{3}\geq\frac{(\sqrt2x+\sqrt2y)^2}{3}=\frac{2}{3}$$ The equality occurs for $(\sqrt2,1)||(x,\sqrt2y)$, which says that $\frac{2}{3}$ is a minimal value.

The maximum does not exist, of course.

Done!

Answer 4

Solution $[1]$
$2y^2+x^2=t⇔\dfrac {y^2}{t/2}+\dfrac{x^2}t=1$

Since this is ellipse, when this is tangent to $x+y=1$, it's minimum. There is no maximum.

$2(1-x)^2+x^2-t=0$

$⇔3x^2-4x+2-t=0$

$D/4=4-3(2-t)=0⇔t=2/3$

Solution $[2]$

$x=\cos^2a,y=\sin^2a$

$t=2y^2+x^2=2\sin^4a+\cos^4a=2\sin^4a+(1-\sin^2a)^2$

$=3\sin^4a-2\sin^2a+1=3(\sin^2a-1/3)^2+2/3\geq\frac23$//

Answer 5

Elementary method

By $y=1-x$, $$2y^2+x^2=2(1-x)^2+x^2=3x^2-4x+2=3(x-\frac23)^2+\frac23$$

which has the minimum value of $\displaystyle\frac23$ at $\displaystyle x=\frac23$.

Answer 6

A bit Geometry:

1) $y + x = 1$.

A straight line with $y-intercept $ $1$, $x- intercept$ $1$, $m = -1$.

2) Consider $2y^2 + x^2 = c^2$, with $c>0$.

Problem: maximize or minimize $c^2$ , given condition 1).

Rewriting :

$\frac{y^2}{(c^2/2)} + \frac{x^2}{c^2} = 1$.

This is an ellipse centered at the origin, with

minor axis $c/√2$ along the $y-axis$, and

major axis $c$ along the $x - axis$.

Smaller $c$ shrinks the ellipse, larger $c$ blows it up.

3 possibilities:

1) The ellipse intersects

$ y + x = 1$ in 2 distinct points.

2) The ellipse touches the given line.

3) The ellipse and line have no point in common.

We are interested in case 2), minimal $c^2$.

2)The tangent to the ellipse has the slope $m = -1$.

Differentiating $2y^2 + x^2 = c^2$ with respect to $x$:

$4y(dy/dx) + 2x = 0$.

$dy/dx = \frac{-x}{2y} = - 1$, or $2y = x$.

Combining with $y + x = 1$ yields

$y = 1/3$, $x = 2/3$.

$min$:

$c^2 = 2y^2 + x^2 $= $2(1/9)+4/9$= $2/3$.