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Exercise 7.4 · Q41

Q.Maximize : z=7x+11yz = 7x + 11y subject to 3x+5y≤26, 5x+3y≤30, x≥0, y≥03x + 5y \le 26,\ 5x + 3y \le 30,\ x \ge 0,\ y \ge 0.

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To maximum z=7x+11yz=7x+11y subject to 3x+5y≤26, 5x+3y≤30, x≥0, y≥03x+5y\le26,\ 5x+3y\le30,\ x\ge0,\ y\ge0: draw each constraint's boundary line, shade the half-plane it demands, and darken the common (feasible) region. Its corner points, found by solving each pair of boundary lines that meet there, are evaluated in the objective function zz (the Corner-Point Theorem guarantees the optimum of a linear objective over a convex polygon occurs at one of its vertices):

Corner point (x,y)(x,y)zz
(0, 0)0
(6, 0)42
(\tfrac{9}{2}, \tfrac52)59
(0, \tfrac{26}{5})57.2

Comparing all the tabulated values, the maximum value is 59, attained at (4.5,2.5)(4.5, 2.5).

✓Final answer

Maximum value of zz = 59, occurring at (4.5,2.5)(4.5, 2.5).

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