A merchant plans to sell two types of personal computers a desktop model and a portable model that will cost Rs. 25000 and Rs. 40000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit, if he does not want to invest more than Rs. 70 lakh and if his profit on the desktop model Rs. 4500 and on portable model is Rs. 5000.
A merchant plans to sell two types of personal computers a desktop model and a portable model that will cost Rs. 25000 and Rs. 40000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit, if he does not want to invest more than Rs. 70 lakh and if his profit on the desktop model Rs. 4500 and on portable model is Rs. 5000.
Let the merchant stocks x desktop computers and y portable computers, We construct the following table:
TypeNumberInvestmentProfitCost per(in Rs.)(in Rs.)computer(in Rs.)Desktopx25000x4500x25000Portabley40000y5000y40000Totalx+y25000x+40000y4500x+5000y
The cost of a desktop model is Rs. 25000 and of a portable model is Rs. 40000.
However, the merhant can invest a maximum of Rs. 70 lakh.
∴ 25000x+40000y≤7000000
So, our problem is to maximize Z = 4500x + 5000y .......(i)
Subject to constraints x+y≤250 ...........(ii)
25000x+40000y≤7000000⇔5x+8y≤1400 ......(iii)
x≤0, y≤0 ..........(iv)
Firstly, draw the graph of the line 5x + 8y = 1400
x0280y1750
Putting (0, 0) in the inequality 5x + 8y ≤ 1400, we have
5×0+8×0≤1400⇒0≤1400 (which is true)
So, the half plane is towards the origin. Since, x, y ≥ 0
So, the feasible region lies in the first quadrant.
Secondly, draw the graph of the line, x + y = 250
x0250y2500
Putting (0, 0) in the inequality x+y≤250, we have 0+0≤250⇒0≤250 (which is true)
So, the half plane is towards the origin.
On solving the equations x+y=250 and 5x+8y=1400, we get B(200, 50).
∴ Feasible region is OABCO.
The corner points of the feasible region are A(250, 0), B(200, 50) and C(0, 175). The values of Z at these points are as follows:
Corner pointZ=4500x+5000yA(250, 0)1125000B(200, 50)1150000→MaximumC(0, 175)875000
The maximum value of Z is 1150000 at point B (200, 50).
Thus, the merchant should stock 200 desktop models and 50 portable models to get the maximum profit of Rs. 1150000.
Let the merchant stocks x desktop computers and y portable computers, We construct the following table:
TypeNumberInvestmentProfitCost per(in Rs.)(in Rs.)computer(in Rs.)Desktopx25000x4500x25000Portabley40000y5000y40000Totalx+y25000x+40000y4500x+5000y
The cost of a desktop model is Rs. 25000 and of a portable model is Rs. 40000.
However, the merhant can invest a maximum of Rs. 70 lakh.
∴ 25000x+40000y≤7000000
So, our problem is to maximize Z = 4500x + 5000y .......(i)
Subject to constraints x+y≤250 ...........(ii)
25000x+40000y≤7000000⇔5x+8y≤1400 ......(iii)
x≤0, y≤0 ..........(iv)
Firstly, draw the graph of the line 5x + 8y = 1400
x0280y1750
Putting (0, 0) in the inequality 5x + 8y ≤ 1400, we have
5×0+8×0≤1400⇒0≤1400 (which is true)
So, the half plane is towards the origin. Since, x, y ≥ 0
So, the feasible region lies in the first quadrant.
Secondly, draw the graph of the line, x + y = 250
x0250y2500
Putting (0, 0) in the inequality x+y≤250, we have 0+0≤250⇒0≤250 (which is true)
So, the half plane is towards the origin.
On solving the equations x+y=250 and 5x+8y=1400, we get B(200, 50).
∴ Feasible region is OABCO.
The corner points of the feasible region are A(250, 0), B(200, 50) and C(0, 175). The values of Z at these points are as follows:
Corner pointZ=4500x+5000yA(250, 0)1125000B(200, 50)1150000→MaximumC(0, 175)875000
The maximum value of Z is 1150000 at point B (200, 50).
Thus, the merchant should stock 200 desktop models and 50 portable models to get the maximum profit of Rs. 1150000.
