WeBWork England MAT x 8/1/?theme &displayMode; Math Jax&user; ckfreela&key; 7NcC

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WeBWork England MAT x 8/1/?theme &displayMode; Math Jax&user; ckfreela&key; 7NcCozoA89xGojt12krqScR https://webwork.asu.edu /webwork2/England MAT 267 Spring 2015 Section Homework Sets Section 11.8 Section 11.8: Problem 1 Problem 1 Password/Email Prev Up Next Grades (1 pt) For each of the following functions, find the maximum and mimimum values of the function on the circular disk: z2 +y2 S 1. Do this by Problems looking at the level curves and gradiants. Problem 1 A) f (a, y y 4 Problem 2 maximum value 2 Problem 3 minimum value 0 Problem 4 Problem 5 B) f(a, y maximum value 2 Display options minimum value 1 View equations as c) f (a, y ages Math Jax maximum value 1 Show saved answers minimum value -2 Yes No Use Equation Editor? Note: You can earn partial credit on this problem. Yes No Preview Answers Submit Answers Apply Options You have attempted this problem 3 times. Your overall recorded score is 0%.

Explanation / Answer

f(x,y) is a plane. To find the critical points we set the partial derivatives (wrt x, y) equal to zero and solve:

fx=1,fy=1, These never equal zero so this function has no critical points and thus no LOCAL max or min values

But, It can still have max or min values at the boundary of the disk described by x^2+y^2=1

The gradient of this function is:

gradf(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).

so by comapring we get   r^2 =1 or r = plus / minus 1

x= r costheta = 1*sqrt2/2   , y= r sin theta = sqrt2/2   

and for r= negative 1

we will have x- -sqrt2/2 and y= -sqrt2/2

so we have coordinates   ( sqrt2/2, sqrt2/2), ( -sqrt2/2/, -sqrt2/2), ( -sqrt2/2/. sqrt2/2 ) ( sqrt2/2, -sqrt2/2)

so we will plug in the given equation f(x, y) =x+y+4   we can easily find maximum and minimum values

maximum we got   4 +sqrt and minimum we got 4 - sqrt2

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