Answer:
<u><em>New houses build =200
</em></u>
<u><em>Profit =$13,860,000</em></u>
Explanation:
In this particular question there are 2 scenarios for demand function,
i.e. (a.) 60 percent chance of low demand, ![P_{60} = 300,000-400Q](https://tex.z-dn.net/?f=P_%7B60%7D%20%20%3D%20300%2C000-400Q)
(b.) 40 percent chance of high demand, ![P_{40} = 500,000-275Q](https://tex.z-dn.net/?f=P_%7B40%7D%20%20%3D%20500%2C000-275Q)
∴ Expected demand function = 60%×(300000-400Q) + 40%×(500000-275Q)
= 380,000-350Q
=380000-350Q
Revenue = (380000 - 350Q)×Q = 380000Q - 350![Q^{2}](https://tex.z-dn.net/?f=Q%5E%7B2%7D)
Here, the no. of new homes build will depend on maximizing profit
∵ Profit = Revenue - Cost
π = (380000 - 350Q)×Q - (140000+240000Q)
π = 380000Q - 350
- (140000+240000Q)
In order to maximize profit , we will
= 0
= 380000-350*2*Q-240000
∴ 380000-350×2×(Q-240000) =0
Q=200
<u>So number of houses that they should build =200
</u>
π = 380000Q - 350
- (140000+240000Q)
π = 380000-(350×
) - (140000+(240000*200))
π =$13,860,000
New houses to be build =200
Profit =$13,860,000