if the price of butternut squash is currently 80 cents per pound, but decreases by 2 cents each day So growing the squash for 10 days will maximize profit
Whenever any product is sold then by selling it the seller earn more then the amount invested in the product. So this earning is called as the profit.
Profit is defined as the
s(t)=120+6t
Now let's say that the price is:
p(t)=0.8-0.02t
Finally we can say that revenue is: (price times supply)
r(t)=[p(t)]*[s(t)]
r(t)=(0.8-0.02t)(120+6t)
r(t)=96+4.8t-2.4t-0.12t^2
r(t)=-0.12t^2+2.4t+96 now that we have the revenue function we take the derivatives...
dr/dt=-0.24t+2.4, d2r/dt2=-0.24
Since the acceleration is a constant negative we can say that when the velocity, dr/dt=0, it is the absolute maximum for r(t)
dr/dt=0 only when 0.24t=2.4, t=10 days
So growing the squash for 10 days will maximize profit. (technically we just used revenue because we had no information as to the cost of growing :P)
If you'd care to check you could simply evaluate r(t) at t=10 and 10±d
r(t)=-0.12t^2+2.4t+96
r(10)=$108,
r(9.99)=$107.999988
r(10.01)=$107.999988
However, the changes in squash and price take place in daily increments and not hundredths of a day, the differences from the maximum profit will be much larger than shown above
r(9)=$107.88, r(11)=$107.88
Hence if the price of butternut squash is currently 80 cents per pound, but decreases by 2 cents each day So growing the squash for 10 days will maximize profit
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