OpenStudy (anonymous):

For each of the following functions, find the maximum and mimimum values of the function on the circular disk: x^2 + y^2 ≤ 1. Do this by looking at the level curves and gradients. (A) f(x, y) = x + y + 3: maximum value = ??? minimum value = ???

6 years ago
OpenStudy (anonymous):

So you are not to use the second derivative test here?

6 years ago
OpenStudy (anonymous):

u can do anything ! it was easier for me when it was like for example 4x^2+3y^2 but when its just simply x or y its confusing !

6 years ago
OpenStudy (anonymous):

i'm gonna go to my other class i'll check the answer there ! thanks for helping out !

6 years ago
OpenStudy (anonymous):

Ok. Well f(x,y) is a plane. To find the critical points we set the partial derivatives (wrt x, y) equal to zero and solve:\[f _{x}=1, f _{y}=1\]These never equal zero so this function has no critical points and thus no LOCAL max or min values. It can still have max or min values at the boundary of the disk described by x^2+y^2=1

6 years ago
OpenStudy (anonymous):

The gradient of this function is:\[grad f(x,y)=<1,1>\]This is the direction of the maximum rate of change of the function. The level curves are just straight lines. Draw the level curves in the x-y plane and draw the boundary (x^2+y^2=1). You should be able to read off this picture the max and min values (which occur at the boundary) of this function.

6 years ago
OpenStudy (anonymous):

So I drew this one out on my whiteboard. I get \[\max=f(1/\sqrt{2},1/\sqrt{2})=3+\frac{2}{\sqrt{2}}\]\[\min=f(-1/\sqrt{2},1/\sqrt{2})=3-\frac{2}{\sqrt{2}}\]

6 years ago
OpenStudy (anonymous):

To see this, start at the origin and move in the direction of the gradient vector <1,1>. At what point do you hit the boundary (unit circle x^2+y^2=1)? The is your max value. To get min value, move in the direction -gradf=<-1,-1>. When you hit the boundary, this is your absolute min of the function.

6 years ago
OpenStudy (anonymous):

good question! let me know if I am correct :)

6 years ago
OpenStudy (anonymous):

it is correct thanks man for helping out !

6 years ago
OpenStudy (anonymous):

one question ! how do u get 1/sqrt2 and similar numbers from that !?

6 years ago
OpenStudy (anonymous):

The max (and min) of the function will occur when the level curves are tangent to the boundary circle. So we want the point at which the slope of the tangent to the circle x^2+y^2=1 is the same as the slope of the tangent line x+y+3=k. All the level curves have the same slope (m=-1) because y=-x-3+k, but different y-intercept values depending on the value of k. So where on the circle does the tangent line have a slope=-1? Differentiating the function x^2+y^2=1 implicitly (and treating y as a function of x) we get:\[2x+2yy'=0\]Solving for y':\[y'=\frac{-x}{y}\]We want this to equal -1:\[-1=\frac{-x}{y}\]Thus:\[x=y\]We put y=x into the equation of the circle and get:\[x^2+x^2=1\]Solving we get:\[x=y=\pm \frac{1}{\sqrt{2}}\]These are the points on the unit circle where the slope of the tangent lines are the same as the slope of the level curves. Another way of saying this is this is these are the points where the level curves just touches the boundary...we can't make the function f(x,y) any bigger (or smaller) than it is at these points.

6 years ago