Find the minimum and maximum values of y in $4x^2 + 12xy + 10y^2 – 4y + 3 = 0$

Question

Find the minimum and maximum values of y in $4x^2 + 12xy + 10y^2 – 4y + 3 = 0.$

$$4x^2 + 12xy + 10y^2 – 4y + 3 = 0$$ $$(2x+3y)^2 +(y-2)^2=1$$ $$y_{max}=3$$ $$y_{min}=1$$

Any other approach for the same.

Answer 1

We need $$\Delta_x\geq0$$ or $$36y^2-4(10y^2-4y+3)\geq0$$ or $$(y-1)(y-3)\leq0$$ or $$1\leq y\leq3,$$which gives the answer because easy to show that the equality occurs in both cases.

Answer 2

Differentiating with respect to $x$ gives $$8x+12y=0$$ We already have $$4x^2 + 12xy + 10y^2 – 4y + 3 = 0 \rightarrow (2x+3y)^2+(y-2)^2=1$$

Both of these must be satisfied in order to find the extrema. Solving for $x$ in the first equality finds $$x = -\frac{3y}{2}$$

Plugging this into the second equation finds $$\left(-3y+3y\right)^2+(y-2)^2=1 \rightarrow (y-2)^2=1 \rightarrow y-2 = \pm 1$$

Finally, we have that $y = 1, 3$ are the minimum and maximum values of $y$ for this equation.

Answer 3

Implicit differentiation of $$4x^2 + 12xy + 10y^2 – 4y + 3 = 0$$

with respect to $x$ implies $$ y'= \frac {-8x-12y}{12x+20y-4}$$

The maximum and minimum points are found by $y'=0$ which implies $$2x+3y=0$$

Since $$4x^2 + 12xy + 10y^2 – 4y + 3 = 0 \implies (2x+3y)^2 +y^2-4y+3=0$$

We get $$y^2-4y +3 =0 $$ that is $y=1$ for the minimum and $y=3$ for the maximum.