We have,

f(x,y)=4x^3+y^2

c(x,y)=2x^2+y^2 with condition c(x,y)=1.

Thus,

grad (f) = k*grad (c)

Thus,

<12x^2,2y> = k * <4x,2y>

Thus,

12x^2 = 4kx

2y = 2yk

2x^2+y^2=1

Simplify them by canceling the constants,

3x^2 = kx (1)

y=yk (2)

2x^2+y^2 = 1 (3)

Look at equation (2),

It tells us:

y=0ork=1.

y=0

Then, (3) tells us that,

2x^2 = 1

Thus, x=+/- sqrt(2)/2

k=1

Then, (1) tells us that,

3x^2 = x

Thus,

x=0orx=1/3.

If x=0 then (3) tells us that y=+/- 1.

If x=1/3 then (3) tells us that y=+/- sqrt(7)/3

Thus, all the critical points are:

(-sqrt(2)/2,0)

(sqrt(2)/2,0)

(0,1)

(0,-1)

(1/3, -sqrt(7)/3)

(1/3), sqrt(7)/3)

Check each one for function:

f(x,y) = 4x^3 +y^2

To see the maximum and minimum.