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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