Find the smallest possible value of the quantity $\displaystyle x^2 + y^2$ under the restriction that $\displaystyle 2x + 3y = 6$

2. Originally Posted by lightningstab714
Find the smallest possible value of the quantity $\displaystyle x^2 + y^2$ under the restriction that $\displaystyle 2x + 3y = 6$

$\displaystyle 2x + 3y = 6 \Rightarrow x = \frac{6 - 3y}{2}$ .... (1)

Substitute (1) into $\displaystyle L = x^2 + y^2$ and find the value of y that minimises L. Substitute this value into L to get the minimum value of L.

3. Thanks.

4. Here are two non calculus approaches to your problem:

Geometrically,$\displaystyle x^2+y^2$ denotes square of distance of the point $\displaystyle (x,y)$ from $\displaystyle (0,0)$ therefore,you have been asked to find the shortest distance between $\displaystyle (0,0)$ and the line $\displaystyle 2x+3y=6$ in 2D space which is the perpendicular distance of 2x +3y=6 from (0,0) and which can be found from the formula
$\displaystyle p=\frac{|ax_{1}+by_{1}+c|}{\sqrt{a^2+b^2}}$
$\displaystyle =\frac{|2(0)+3(0)+6|}{\sqrt{2^2+3^2}}$
$\displaystyle =\frac{6}{\sqrt{13}}$.Therefore answer should be $\displaystyle \frac{36}{13}$

Another method:

Put $\displaystyle x=r\cos\theta;y=r\sin \theta$ where $\displaystyle r=\sqrt{x^2+y^2}$

Plugging in $\displaystyle 2x+3y=6$,we get

$\displaystyle r=\frac{6}{2\cos \theta +3\sin \theta}$

Now r will be minimum when denominator will be maximum.

Also$\displaystyle ,-\sqrt{a^2+b^2} \leq a\cos\theta+b\sin\theta \leq\sqrt{a^2+b^2}$

Therefore,$\displaystyle r_{min}=\frac{6}{\sqrt{2^2+3^2}}=\frac{6}{\sqrt{13 }}$
and hence $\displaystyle x^2+y^2=\frac{36}{13}$