# Thread: Hard minimum value problem

1. ## Hard minimum value problem

Show that if x and y are real numbers and $\displaystyle 7x^2+3xy+3y^2=1$ then the least positive value of $\displaystyle \frac{x^2+y^2}{y}$ is 0.5.

I've tried lots of methods but none work.

2. ## Ok

[quote=0-);131057]Show that if x and y are real numbers and $\displaystyle 7x^2+3xy+3y^2=1$ then the least positive value of $\displaystyle \frac{x^2+y^2}{y}$ is 0.5.

Solve the first for either x or y...imput that into the second formula...differentiate and find the minimum...then to get the other value imput your one value into the original equation and solve for the other value...and to make sure you are getting a minimum plug your answer into the second derivative and make sure its positive

3. [quote=Mathstud28;131089]
Originally Posted by 0-)
Show that if x and y are real numbers and $\displaystyle 7x^2+3xy+3y^2=1$ then the least positive value of $\displaystyle \frac{x^2+y^2}{y}$ is 0.5.

Solve the first for either x or y...imput that into the second formula...differentiate and find the minimum...then to get the other value imput your one value into the original equation and solve for the other value...and to make sure you are getting a minimum plug your answer into the second derivative and make sure its positive
Thank you.

I thought there would be a quicker way of doing it but your way should work.