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3. Find the absolute maximum and minimum values of the function f(x,y)=x-2.xy +2y on the rectangle D = {(x, y)|0

3. Find the absolute maximum and minimum values of the function f(x,y)=x-2.xy +2y on the rectangle D = {(x, y)|0

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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(3. Find the absolute maximum and minimum values of the function ƒ(x,y)=x² -2.xy+2y on the rectangle D= {(x, y)|0<x<3,0< y<2}. 3. Since fis a polynomial, It is continuous on the closed, bounded rectangle D. so there is both an absolute maximum and an absolute minimum. we first find the critical points. These occur when f, = 2x - 2y= 0 f, = -2x + 2 =0 so the only critical point is (1, 1), and the value of f there is (1, 1) = 1. we look at the values of F on the boundary of D, which consists of the four line segments L, L, L, L. (2,2) (0, 2) (3, 2) L. (0,0) (3,0) On L, we have y=0 and f(x, 0) = x This is an increasing function or x. so 1ES minimum value is rtu. 0) = 0 and its maxi- mum value is F(3, 0) = 9. On Lz we have r= 3 and (3. y) = 9 - 4y 0 <y 2 This is a decreasing function of y, so its mmaximum value is f(3, 0) = 9 and its minimum value is f(3, 2) = 1. On La we have y= 2 and f(x, 2) = x-4x + 4 0 <x3 or simply by observing that f(x, 2) = (x- 2), we see By the methods of Cal. 1. that the minimum value of this function is f(2, 2) = 0 and the maximum value is (0, 2) = 4. Finally, on L we have x= 0 and r(0. y) = 2y 0 y 2 with maximum value f(0, 2) = 4 and minimIum value N0. 0) = 0. Thus, on the bound- ary, the minimum value of f is 0 and the maximum is 9. we compare these values with the value f(1, 1) = 1 at the critical point and conclude that the absolute maximum value of f on Dis 3, 0) = 9 and the absolute minimum value is (0, 0) = F(2, 2) = 0. Figure 13 shows the graph of f.
Transcribed Image Text:(3. Find the absolute maximum and minimum values of the function ƒ(x,y)=x² -2.xy+2y on the rectangle D= {(x, y)|0<x<3,0< y<2}. 3. Since fis a polynomial, It is continuous on the closed, bounded rectangle D. so there is both an absolute maximum and an absolute minimum. we first find the critical points. These occur when f, = 2x - 2y= 0 f, = -2x + 2 =0 so the only critical point is (1, 1), and the value of f there is (1, 1) = 1. we look at the values of F on the boundary of D, which consists of the four line segments L, L, L, L. (2,2) (0, 2) (3, 2) L. (0,0) (3,0) On L, we have y=0 and f(x, 0) = x This is an increasing function or x. so 1ES minimum value is rtu. 0) = 0 and its maxi- mum value is F(3, 0) = 9. On Lz we have r= 3 and (3. y) = 9 - 4y 0 <y 2 This is a decreasing function of y, so its mmaximum value is f(3, 0) = 9 and its minimum value is f(3, 2) = 1. On La we have y= 2 and f(x, 2) = x-4x + 4 0 <x3 or simply by observing that f(x, 2) = (x- 2), we see By the methods of Cal. 1. that the minimum value of this function is f(2, 2) = 0 and the maximum value is (0, 2) = 4. Finally, on L we have x= 0 and r(0. y) = 2y 0 y 2 with maximum value f(0, 2) = 4 and minimIum value N0. 0) = 0. Thus, on the bound- ary, the minimum value of f is 0 and the maximum is 9. we compare these values with the value f(1, 1) = 1 at the critical point and conclude that the absolute maximum value of f on Dis 3, 0) = 9 and the absolute minimum value is (0, 0) = F(2, 2) = 0. Figure 13 shows the graph of f.
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