A company produces two types of gadgets: Alpha and Beta. Each Alpha requires 3 hours of assembly and 2 hours of testing. Each Beta requires 2 hours of assembly and 4 hours of testing. The company has at most 240 hours of assembly and 320 hours of testing available per week. If the profit per Alpha is $50 and per Beta is $60, and the company must produce at least 20 Alpha and at least 30 Beta, what is the maximum weekly profit?

Jawaban: $4,000

Let A = number of Alpha, B = number of Beta. Constraints: 3A + 2B ≤ 240, 2A + 4B ≤ 320, A ≥ 20, B ≥ 30. Profit P = 50A + 60B. Test vertices: (20,30): P=1000+1800=2800. (20,70): check 2(20)+4(70)=40+280=320 OK; 3(20)+2(70)=60+140=200 ≤240. P=1000+4200=5200. Wait, that's higher. But 70 Beta: assembly 200, testing 320. Profit 5200. Next (60,30): 3(60)+2(30)=180+60=240; 2(60)+4(30)=120+120=240 ≤320. P=3000+1800=4800. Next intersection of 3A+2B=240 and 2A+4B=320. Solve: multiply first by 2: 6A+4B=480; subtract second: 4A=160 → A=40; then 3(40)+2B=240 → 120+2B=240 → B=60. P=50(40)+60(60)=2000+360