Find the maximum and minimum values attained by the function f along the path c(t).
f(x, y) = xy; c(t) = (cos(t), sin(t)); 0 ≤ t ≤ 2π maximum value minimum value.
f(x, y) = x² + y²; c(t) = (cos(t), 4 sin(t)); 0 ≤ t ≤ 2π maximum value minimum value.

Solution:

We have to find the maximum and minimum values attained by the function f along the path c(t).

a) f(x, y) = xy; c(t) = (cos(t), sin(t)); 0 ≤ t ≤ 2π

f(x, y) = xy

f(x, y) = cos t . sin t

By using trigonometric identity,

sin(2x) = 2 sin(x) cos(x)

So, sin(x) cos(x) = (1/2)sin(2x)

Now, f(x,y) = (1/2)sin(2t)

We know that, -1 ≤ (1/2)sin(2t)  ≤ 1

(-1/2) ≤ (1/2)sin(2t)  ≤ (1/2)

(-1/2) ≤ f(x, y) ≤ (1/2)

Therefore, the minimum value is -1/2 and the maximum value is 1/2.

b) f(x, y) = x² + y²; c(t) = (cos(t), 4 sin(t)); 0 ≤ t ≤ 2π

f(x, y) =  x² + y²

f(x, y) =  (cos t)² + (4sin t)²

f(x, y) = cos²t + 16sin²t

We know, cos²x + sin²x = 1

So, f(x, y) = cos²t + sin²t + 15sin²t

f(x, y) = 1 + 15sin²t

We know that,

0 ≤ sin²t ≤ 1

0 ≤ 15sin²t ≤ 15

1 ≤ 1 + 15sin²t ≤ 16

1 ≤ f(x, y) ≤ 16

Therefore, the minimum value is 1 and maximum value is 16.

Therefore, 

• f(x, y) = xy; c(t) = (cos(t), sin(t)); 0 ≤ t ≤ 2π maximum value is 1/2 and  minimum value is -1/2.
• f(x, y) = x² + y²; c(t) = (cos(t), 4 sin(t)); 0 ≤ t ≤ 2π maximum value is 16 and minimum value is 1.

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Find the maximum and minimum values attained by the function f along the path c(t).

Summary:

The maximum and minimum values attained by the function f along the path c(t). f(x, y) = xy; c(t) = (cos(t), sin(t)); 0 ≤ t ≤ 2π maximum value is 1/2 and  minimum value is -1/2. f(x, y) = x² + y²; c(t) = (cos(t), 4 sin(t)); 0 ≤ t ≤ 2π maximum value is 16 and minimum value is 1.