The figure below shows some of the contours of f(x, y) on a portion of the xy-pl

Question

The figure below shows some of the contours of f(x, y) on a portion of the xy-plane. Remember, there are an infinite number of contours. (a) What is the maximum value of f(x, y) on the rectangle 20 lessthanorequalto x lessthanorequalto 30, 0 lessthanorequalto y lessthanorequalto 5? (b) Let B be the rectangle 0 lessthanorequalto x lessthanorequalto 5. Use the maximum value of f(x, y) on B, to find an overestimate for integral_B f(x, y) dA. (c) If R is the rectangle 0 lessthanorequalto x lessthanorequalto 30, 0 lessthanorequalto y lessthanorequalto 10 and we use Delta x = 10, delta y = 5, how many rectangular boxes should we use to approximate integral_R f(x, y) dA? (d) Use the maximum values of f(x, y) on each subrectangle in part (c) to find an overestimate for integral_R f(x, y) dA.

Explanation / Answer

a) The maximum value in the rectangle is 8. You can see that the countour with value 8 is touching the rectangle at the corner (30,5).

b) The maximum value of the function in the rectangle B is 4. Being a constant you can take this maximum outside the integral. Therefore the given integral < f_max * area of the rectangle = 4*(10-0)*(5-0)=200. That is 200 is an upperbound or overestimate for the given integral.

c) Since Delta x=10 and Delta y=5, we need 6 boxes to approximate the givven integral over the rectangle R.

[since Total area / area of rectangular sub box = ((30-0)*(10-0)) / (Delta x* Delta y)=(30*10) / (10*5) = 30 / 5 = 6. ]

d) Area of a sub rectangle = Delta x*Delta y = 50. The given integral can be bounded by summation 50*f_max, over the 6 subrectangles. Since f_max is 4,6,6,8,8,10 respectively, we see that integral is bounded by

50*4+50*6+50*6+50*8+50*8+50*10 = 50*(4+6+6+8+8+10) = 50*42=2100. Hence 2100 is the over estimate for the given integral.

Thumbs up :)

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