A fruit grower knows from previous experience and careful data analysis that if the fruit on a specific kind of tree is harvested at this time of year, each tree will yield, on average, 137 pounds, and will sell for $2.4 per pound. However, for each additional week the harvest is delayed (up to a point), the yield per tree will increase by 2.5 pounds, while the price per pound will decrease by $0.04.

A) t = 2.6 weeks

B) R = $329.48

Step-by-step explanation:

At the right time, the average amount of money made from the fruits is given as

R = (selling price of fruits per pound) × (average number of pounds of fruits) = 2.4 × 137 = $328.8 on average.

So, delaying the harvest by a week makes the revenue turn to

R = (2.4 - 0.04)(137 + 2.5)

Delaying the harvest by t weeks makes the revenue turn to

R = (2.4 - 0.04t) × (137 + 2.5t)

R = 328.8 + 6t - 5.48t - 0.1t²

R(t) = -0.1t² + 0.52t + 328.8

A) Time in weeks when the Revenue would be highest.

That is, value of t that corresponds to the maximum value of R(t)

At maximum value, (dR/dt) = 0 and (d²R/dt²) < 0, that is negative.

R(t) = -0.1t² + 0.52t + 328.8

(dR/dt) = -0.2t + 0.52 = 0

and (d²R/dt²) = -0.2 < 0 (negative; this is indeed a maximum point for the function)

When (dR/dt) = 0,

(dR/dt) = -0.2t + 0.52 = 0

-0.2t + 0.52 = 0

t = (0.52/0.2)

t = 2.6 weeks.

B) Actual Maximum value of the Revenue.

This corresponds to the revenue when t = 2.6 weeks

R(t) = -0.1t² + 0.52t + 328.8

R(2.6) = -0.1(2.6)² + 0.52(2.6) + 328.8

R(t = 2.6) = -0.676 + 1.352 + 328.8

R at maximum point = $329.48

Hope this Helps!!!

the answer is b; false. 2 dimensional things are a lot more different from 3 dimensional things.

i hope this ! : d

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