2024/035) What are the maximum and minimum values of 3x+4y on the circle $x^2+y^2=1$

As $x^2 + y^ 2 = 1$ we can chose $x = \sin\, t , y = \cos\, t$

$3x + 4 y= 3 ]sin\, t + 4 \cos\, t$

To convert $3x + 4 y= 3 ]sin\, t + 4 \cos\, t$ to the form $A \sin (x+ t)$

$A \sin (x+t) = A \sin\,t \cos\, x + A \cos\, t \sin\, x$

We can choose $3 = 5 \cos\, x$ and $4 = 5\ sin\, x$ (as $3^2 + 4^2 = 25 = 5^2$)

$= 5 \cos\, x \sin\, t + 5 \cos\, t \sin\, x = 5 \sin (x-t)$

It is maximum when $\sin (x-t) = 1$ and maximum value = $5$

minimum when $\sin (x-t) = -1$ and minimum value = $-5$