Question

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

Answer 1

Answer:Check the attached

Step-by-step explanation:

Answer 2

Answer:

a) Therefore, the minimum value of the function f is -1/2.

Therefore, the maximum value of the function f is 1/2.

b)Therefore, the minimum value of the function f is 1.

Therefore, the maximum value of the function f is 16.

Step-by-step explanation:

We find the maximum and minimum values attained by the function f along the path c(t). From exercise we have:

a)

$f(x,y)=xy\\\\c(t)=(\cos t, \sin t)\\\\\0\leq t\leq 2\pi$  

Therefore, we get:

$f(x,y)=xy\\\\f(x,y)=\cos t \cdot \sin t\\\\f(x,y)=\frac{1}{2} \sin 2t$

We know that:

$-1\leq \sin 2t\leq 1\\\\\implies -\frac{1}{2} \leq \frac{1}{2}\sin 2t\leq \frac{1}{2}\\\\\implies -\frac{1}{2}\leq f(x, y)\leq \frac{1}{2}$

Therefore, the minimum value of the function f is -1/2.

Therefore, the maximum value of the function f is 1/2.

We use software to drawn a graph.

b)

$f(x,y)=x^2+y^2\\\\c(t)=(\cos t, 4\sin t)\\\\0\leq t\leq 2\pi$

Therefore, we get:

$f(x,y)=x^2+y^2\\\\f(x,y)=(\cos t)^2 \cdot (4\sin t)^2\\\\f(x,y)=\cos^2t+16\sin^2t\\\\f(x,y)=\cos^2t+\sin^2t+15\sin^2t\\\\f(x,y)=1+15\sin^2t$

We know that:

$0\leq \sin^2t\leq 1\\\\\implies 0\leq 15\sin^2t\leq 15\\\\1\leq 1+15\sin^2t\leq 16\\\\ \implies 1\leq f(x,y) \leq 16$

Therefore, the minimum value of the function f is 1.

Therefore, the maximum value of the function f is 16.

We use software to drawn a graph.