Find the maximum and minimum values of f(x,y)=6x+y on the ellipse x^2+49y^2=1

Answer 1

ellipse has center at the origin top vertex (0,1/7), right vertex (1,0), bottom vertex (0,-1/7)

left vertex (-1,0) the maximum of f(x,y) on the ellipse is near the right vertex

the minimum is near the left vertex

the top vertex is the max point on the ellipse

the bottom vertex is the min point on the ellipse

both max and min are close to y=0

it might help to sketch a rough graph of the ellipse and lines

6x-y=f

f'= y'= -6

x^2+49y^2=1

49y^2=1-x^2

y=+/-(sqr(1-x^2)/7

y'= +/-(1/7)(1/2)(-2x)/sqr(1-x^2)

=+/-x/7sqr(1-x^2)=-6

x=42sqr(1-x^2)

x^2=1764-1764x^2

1765x^2=1764

x^2=1764/1765

x=+/- sqr(1764/1765)

y=+/-(sqr(1-1764/1765)/7

= +/-1/7sqr1765

=about 0.0034 = max f(x,y)

= about -0.0034 = min f(x,y)

where (x,y) is on the ellipse and f(x,y) is tangent to the ellipse