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Solving problems in higher mathematics in freshman year.
The detailed process is that there is a price P=5-x/200 under the condition of topic setting.

Also, ∫ total cost C= fixed cost+variable cost =C(0)+ output × unit variable cost =5+4x, total sales revenue R= sales volume × unit selling price =x*P=x(5-x/200),

However, sales profit Prof(x)= total sales revenue R- total cost C=x(5-x/200)-(5+4x)=x-x? /200-5。

Derive x from Prof(x) so that its value is 0, and prof' (x) =1-x/100 = 0. ∴x= 100。

Obviously, there is a maximum value of Prof(x), and the extreme point x= 100 is unique. Therefore, when x= 100, that is, the production and sales volume is 100, the profit is the largest, and its value is 45 (ten thousand yuan). At this time, the price P=5- 100/200=4.5 (ten thousand yuan/100 sets). ∴ Price per set p=450 yuan.

For reference.