Let the seller increases Rs x in annual subscription.
∴ Total revenue generated by her, R(x)= Rs (300+x)(500−x)
Diff. w.r.t. x both sides, R′(x)=−(300+x)+(500−x)=200−2x
Again diff. w.r.t. x,R"(x)=−2
For the points of local maxima & local minima, R′(x)=0⇒x=100
Now, R"(x=100)=−2<0 ∴R(x) is maximum at x=100.
So, an increment of Rs 100 will maximize the profit.
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