(a) A factory manufactures two types of screws, A and B. Each type of screw requires the use of two machines, an automatic and a hand operated. It takes 4 minutes on the automatic and 6 minutes on hand operated machines to manufacturer a package of screw A, while it takes 6 minutes on automatic and 3 minutes on the hand operatedmachines to manufacture a package of screw B. Each machine is available for at the most 4 hours (240minutes) on any day. The manufacturer can sell a package of screws A at a profit of Rs 7 and screws B at a profit of Rs.10. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximize his profit? Determine the maximum profit.
(b)Prove that∣∣ ∣∣1+a1111+b1111+c∣∣ ∣∣=abc(1+1a+1b+1c)=abc+bc+ca+ab
Let x, y be number of packs of
(a) Screw A and Screw B respectively
⇒z=7x+10y for graph
x≥0,y≥0 and 4x+6y≤240 for corner point .
6x+3y≤240
Using corner point method the maximum profit occurs at x = 30, y = 20
⇒ maximum profit = Rs 410
∴∫2a0f(x)dx=∫a0f(x)dx+∫a0f(2a−x)dx={2∫a0f(x)dx,if f(2a−x)=f(x)0,if f(2a−x)=−f(x)I=∫a0f(x)dxf(x)=cos5x⇒f(2π−x)=cos5(2π−x)=cos5xI=2∫a0cos5xdxf(π−x)=cos5(π−x)=−cos5x=−f(x)⇒I=0
(b)∣∣ ∣∣1+a1111+b1111+c∣∣ ∣∣=abc∣∣ ∣ ∣ ∣∣1+1a1b1c1a1+1b1c1a1b1+1c∣∣ ∣ ∣ ∣∣C1→C1+C2+C3=abc∣∣ ∣ ∣ ∣∣1+1a+1b+1c1b1c1+1a+1b+1c1+1b1c1+1a+1b+1c1b1+1c∣∣ ∣ ∣ ∣∣
=abc(1+1a+1b+1c)∣∣ ∣ ∣ ∣∣11b1c11+1b1c11b1+1c∣∣ ∣ ∣ ∣∣R2−R1,R3−R1=abc(1+1a+1b+1c)∣∣ ∣ ∣∣11b1c010001∣∣ ∣ ∣∣=abc(1a+1b+1c+1)=abc+ab+bc+ca