Question

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

Answer 1

$4x^2+12xy+10y^4-4y+3=0$

On differentiating, we get

$\Rightarrow 8x+12x\frac{dy}{dx}+12y+40y^3\frac{dy}{dx}-4\frac{dy}{dx}=0$

for $max/min,\frac{dy}{dx}=0$

so, $8x+12y=0$

$x=\frac{-3}{2}y$

Put in

$\Rightarrow f(x,y)=4x^2+12xy+10y^4-4y+3$

$max\left\{f(x,y)\right\}=9y^2-18y^2+10y^4-4y+3$

$max\left\{f(x,y)\right\}=10y^4-9y^2-4y+3$

which occurs at $x=\frac{-3}{2}y.$

Hence, solved.