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.

"> 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.

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The profit function, in rupees, of a firm selling x items \[(x\ge 0)\] per week is given by\[P(x)=-3500+(400-x)x\]. How many items should the firm sell so that the firm has maximum profit?

Joint Entrance Exam - Main (JEE Main) Mathematics in Joint Entrance Exam - Main (JEE Main) 10 months ago

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[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.

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Posted on 18 Aug 2024, this text provides information on Joint Entrance Exam - Main (JEE Main) related to Mathematics in Joint Entrance Exam - Main (JEE Main). Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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