A) 400
B) 300
C) 200
D) 100
Correct Answer: C
Solution :
[c] \[P(x)=-3500+(400-x)x=\] \[-3500+400x-{{x}^{2}}\] On differentiating w.r.t.x, we get \[P'(x)=400-2x\] Put \[P'(x)=0\] for maxima or minima \[\Rightarrow 400-2x=0\] \[\Rightarrow x=200\] Now \[P''(x)=-2x\] \[\Rightarrow P''(200)=-400<0\] \[\therefore \,\,\,\,\,P(x)\] is maximum at \[x=200\] Hence 200 items should the firm sell so that the firm has maximum profit.You need to login to perform this action.
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