Consider the feasible set in fig 11 a find an objective func

Consider the feasible set in fig 11 (a), find an objective function of the from ax +by that has its greatest valve at given point.

Solution

We know that the maximum distance of a point on a circle is from the center. We will use this property to find the desired maximum distance of the point ( 8, 3) from a point ( h, k) ( say) which is the center of th circle passing through the points ( 9,0), ( 3,8), * 8, 3) and ( 9, 0). Further, if the point ( h, k) is in the feasible region, then we must have h 9 and k 9.

The equation of a circle with radius r and the center ( h, k) is ( x - h)² + ( y - k)² = r^{2 .} If this circle passes through the points (0,9) ), then (0 -h)² + ( 9 -k)² = r² or h² + k² - 18k + 81 = r² or, h² + k² - r² = 18k - 81.

Similarly, if this circle passes through the points (9, 0) ), then (9 -h)² + ( 0 -k)² = r² or h² + k² - 18h + 81 = r²    or, h² + k² - r² = 18h - 81. Therefore,  18h - 81=  18k - 81 or, h = k. Then the equation of the circle changes to ( x - h)² + ( y - h)² = r²....(1) Now, if this circle passes through the point ( 3, 8), then ( 3 - h)² + ( 8 - h)² = r² or, 9 - 6h + h² + 64 - 16h + h² = r² or, 2h² - 22h + 73 = r² ....(2) We can easily verify that this circle passes through the point ( 8,3) also by substituting x = 8 and y = 3 in the equation of the circle which will give us the same equation as equation (2).

Now, we have to find a value for h in the feasible region so that r² ( and hence r) is maximum. For this, we must have dr²/ dh = 0 or,4h-22 = 0 so that h = 22/4 = 11/2. We may observe that since 11/2 <  9, the point ( 11/2, 11/2) is in the feasible region. Then r² = 2h² - 22h + 73 = 2 (11/2)² -22(11/2) + 73 = 2* 121/4 - 121 + 73 = 73 - 121/2 = (146 -121)/2 = 25/2 . Hence r = 5/2.

The equation of a line is y = mx + c where m is the slope.Since this line passes through the point (8,3), and ( 11/2,11/2), therefore, m = ( 3 – 11/2) /(8-11/2) =( -5/2)/ (5/2) = -1. Then, we have y = -x + c . Now, on substituting x = 8 and y = 3, we have 3 = -8 + c so that c = 11. Hence the desired equation is y = -x + 11 or, x + y – 11 = 0. The segment of this line between the points ( 11/2, 11/2) and ( 8,3) has the maximum value at the point ( 8,3).