brief calculus problem | Mathematics | Page 1

jovial cargo Mar 28, 2026, 9:38 PM

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i tried entering the quanitity into the c prime function 700q^-.3-130, why doesn't that work. I get like 0 or 2.4341483E^-4

glass inletBOT Mar 28, 2026, 9:38 PM

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jovial cargo Mar 28, 2026, 9:39 PM

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my original set up was q(1000/q^.3)-130q

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i also had a question of why its not q(1000/q^.3)-130(1000/q^.3)

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profit functions for me are tricky cause i believe it can be either?

vivid lagoon Mar 28, 2026, 9:44 PM

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so you're still on part (a) right ?

jovial cargo Mar 28, 2026, 9:45 PM

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part b is different?

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i thought to find the maximum profit for 1 i would just put it into the derivative equation but it's not working

vivid lagoon Mar 28, 2026, 9:45 PM

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I'm just not sure where you're stuck at

jovial cargo Mar 28, 2026, 9:45 PM

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c

vivid lagoon Mar 28, 2026, 9:45 PM

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ok so you're still on the max profit question

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so part (a)

vivid lagoon Mar 28, 2026, 9:46 PM

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jovial cargo my original set up was q(1000/q^.3)-130q

this should be the total profit yes

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q*p is how much you make from raw sales

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and you lose 130 per unit, so -130q

upbeat horizon Mar 31, 2026, 4:16 AM

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i also had a question of why its not q(1000/q^.3)-130(1000/q^.3)
The first part, q(1000/q^.3), represents the total revenue correctly.
The second part should represent the total cost. For that, we need unit cost times quantity, -130q.
However, 130 * (1000 / q^.3) = 130 * p is actually cost times price, not cost times quantity.

#brief calculus problem