Find the maximum and minimum values of f(x,y)=7x+y on the ellipse x2+9y2=1?

Please use ^ or ² for squared! Here's a few more symbols you can copy and use:
√π∛÷±≠θ⁰¹²³⁴⁵⁶⁷⁸⁹∠∆∟⊥≤≥∴∞

We can define the ellipse parametrically as:
x = cos(t), y = 1/9 sin(t)

Then, the values along the ellipse are:
ƒ = 7x + y
ƒ = 7cos(t) + 1/9 sin(t)
dƒ/dt = -7sin(t) + 1/9 cos(t)

Let dƒ/dt = 0 for extreme points:
7sin(t) = 1/9 cos(t)
t = arctan(1/63), arctan(1/63) + π

Putting that back into the ellipse:
ƒ = 7cos(t) + 1/9 sin(t)
ƒ = 7cos(arctan(1/63)) + 1/9 sin(arctan(1/63))
ƒ = 411 / √[3970] + 1 / 9√[3970]
ƒ = √[3970] / 9

If you use the other t:
ƒ = -√[3970] / 9

Do you know what the answer is?