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I need to find the smallest value of $x^2 + y^2$ with the restriction $2x + 3y = 6$. This chapter focuses on the vertex formula.

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What's the "vertex formula"? Are you talking about this? – Srivatsan Nov 8 '11 at 16:15
@Srivatsan yes, -b/2a – erimar77 Nov 8 '11 at 16:23
up vote 5 down vote accepted

Rewrite $x^2 + y^2$ in terms of one of the variables (either $x$ or $y$) using the restriction given to you which will give you a quadratic equation. That should help move you along to the answer.

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Using the Cauchy-Schwarz inequality, we have $6^2=(2x+3y)^2\leq (x^2+y^2)(2^2+3^2)$, which gives the minimum value of $x^2+y^2$ to be $\frac{36}{13}$.

Edit:Equality occurs for $\frac{x}{2}=\frac{y}{3}$.

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This is great!+1 – Tapu Nov 8 '11 at 18:43
Thank you very much. – Eisen Nov 8 '11 at 18:46

For fun, we give a couple of solutions that are not the intended ones. The solutions are very similar, but the first is expressed algebraically, while the second brings in the geometry.

$1$) Note that $$(2x+3y)^2+(3x-2y)^2=13(x^2+y^2).$$ Thus, given that $2x+3y=6$, $$13(x^2+y^2)= 36+(3x-2y)^2.$$ If we can manage to make $3x-2y=0$, then $13(x^2+y^2)$ will be as small as possible. But the system of two linear equations $2x+3y=6$, $3x-2y=0$ has a solution. There, $$13(x^2+y^2)=36,$$ so the smallest possible value of $x^2+y^2$ is $36/13$.

$2$) Look at the problem geometrically. We want to find the smallest radius $r$ such that the circle $x^2+y^2=r^2$ meets the line $2x+3y=6$. If we draw a picture, we can see that for this smallest $r$, the line $2x+3y=6$ will be tangent to the circle. Let the point of tangency be $T(a,b)$. The line from the origin to $T$ is perpendicular to the tangent line.

The line $2x+3y=6$ has slope $-2/3$. So the line from the origin to $T$ has slope the negative of the reciprocal of $-2/3$. Thus $$\frac{b}{a}=\frac{3}{2}.$$ This equation simplifies to $3a-2b=0$. We also have $2a+3b=6$. Now we can solve for $a$ and $b$. But let's not bother. Use the fact that $$(3a-2b)^2+(2a+3b)^2=13(a^2+b^2)$$ to conclude that $r^2=a^2+b^2=36/13$.

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Let's solve for $y$ in the equation $2x + 3y = 6$. This gives $y = 2 - \frac{2x}{3}$.

After rewriting $x^2 + y^2$ in terms of $x$, we have

$x^2 + y^2 = x^2 + (2 - \frac{2x}{3})^2 = x^2 + (4 - 2 \cdot \frac{4x}{3} +\frac{4x^2}{9})$

This is a quadratic "in $x$" that we would like to minimize. You can write it in standard form $ax^2 + bx + c$ and then use the methods you have learned.

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While this particular problem can be solved more simply, it is worthwhile to solve the problem through Lagrange mulutipliers, which is less ad hoc, if slightly more involved.

Let $f(x,y)=x^2+y^2$ and $g(x,y)=2x+3y-6$. Then a minimum of $f(x,y)$ subject to the constraint $g(x,y)=0$ must satisfy $\nabla f = \lambda \nabla g$ for some real number $\lambda$. Writing vectors in coordinates, this yields

$$(2x,2y)=(2 \lambda, 3 \lambda)$$

or, $\lambda=x=2/3 y$, and the system of equations $x=2/3y$ and $2x+3y=6$ can be solved to determine, $x,y$ and $x^2+y^2$

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