Maximize the function?

Question

Maximize the function f(x,y) = 3x-y+1 on the ellipse x^2 + 3y^2 = 16 .

Please explain what you are doing so that I can learn this stuff. Thanks.

Answer 1

f(x,y) = 3x - y + 1 . . . . . . . equation 1
x^2 + 3 y^2 = 16
y = sqr [ ( 16 - x^2 ) / 3 ] . . . equation2 . . substitute to 1
f(x,y) = ( 3x - sqr [ ( 16 - x^2 ) / 3 ] + 1 . . . . differentiate this
f(x,y) = [ 3 + 2/3 x ] / ( 2 sqr [ ( 16 - x^2 ) / 3 ] ) = 0
3 + 2/3 x = 2 sqr [ ( 16 - x^2 ) / 3 ] . . . squaring both sides
9 + 4x + 4/9 x^2 = 4/3 ( 16 - x^2) . . . .. simplifying
16x^2 + 36x - 111 = 0
x = - 3.98911 . . . & . . x = 1.73911
y = 0.1703 . . . . .& . . y = 2.08

Answer 2

Well, you should first explain what you mean by maximizing 3x-y+1 on the ellipse x^2 +3y^2 = 16.

The ellipse in standard notation is x^2/16 + y^2/(16/3) = 1
So a = 4 and b= (4/3)sqrt(3). Thus c= sqrt(a^2-b^2).
Thus c = sqrt(16 - 16/3) = (4/3)sqrt(6).
Thus the vertices of the ellipse are at (-4,0) and (4,0)
The foci are at (-4/3sqrt(6),0) and (4/3sqrt(6),0)
The major axis = 2a and the minor axis = 2b.

Now y= 3x-1 will be max when x is max
Max value of x is 4 so y = 3x-1 = 12-1 =11 = max of f(x,y).

I don't know if this is what you meant. If not, just ignore it.