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• Use at least twice as many radio advertisements as television advertisements: y ≥ 2 x → 2 x y ≤ 0 • Use no more than 20 television advertisements x ≤ 20 • The television budget should be at least \$140,000 10,000 x ≥ 140,000 → x ≥ 14 • The radio advertising budget is restricted to a maximum of \$99,000 3,000 y ≤ 99,000 → y ≤ 33 • The newspaper budget is to be at least \$30,000 1,000 z ≥ 30,000 → z ≥ 30 The complete LP problem is then: Maximize P = 90*min( x , 10) + 55*max(0, x – 10) + 25*min( y , 15) + 20*max(0, y – 15) + 10*min( z , 20) + 5*max(0, z – 20) Subject to: 40*min( x , 10) + 15*max(0, x – 10) + 20*min( y , 15) + 12*max(0, y – 15) + 10*min( z , 20) + 8*max(0, z – 20) ≥ 1,000
Module B 10 x + 3 y + z ≤ 279 2 x y ≤ 0 x ≥ 14 x ≤ 20 y ≤ 33 z ≥ 30 with x , y , z ≥ 0 For part 2, the maximum budget is changed to 289,000, and the complete LP problem becomes: Maximize P = 90*min( x , 10) + 55*max(0, x – 10) + 25*min( y , 15) + 20*max(0, y – 15) + 10*min( z , 20) + 5*max(0, z – 20) Subject to: 40*min( x , 10) + 15*max(0, x – 10) + 20*min( y , 15) + 12*max(0, y – 15) + 10*min( z , 20) + 8*max(0, z – 20) ≥ 1,000 10 x + 3 y + z ≤ 289 2 x y ≤ 0 x ≥ 14 x ≤ 20 y ≤ 33 z ≥ 30 with x , y , z ≥ 0
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