The Apple Deliverer

Calculus Level 3

An apple deliverer sells his apples to apple shops. From the place where he obtains the apples, there is a shop after every mile. One day he forgets to bring his meal with him and he has to eat the apples for hunger. He has 100 apples and his hunger increases by one apple every mile; that is, he eats N N apples at the N th N^{\text{th}} mile.

If every shop gives him $5 more per apple than the previous shop (and first shop will pay $5 per apple) , what shop provides him maximum profit if he wants to sell all of his apples only at one shop? (On each mile, he first eats the apples and then he can sell the remaining apples to the shop.)


The answer is 8.

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3 solutions

Nicolas Bryenton
Mar 2, 2014

The number of apples eaten per mile is equal to the number of miles traveled. So, the total number of apples eaten at any given mile would be the sum of all the apples eaten before as well as those eaten at that mile. e.g. 1 + 2 + 3 + 4 = 10 1+2+3+4=10 apples eaten after 4 miles This follows the triangular numbers series, which is represented as n 2 + n 2 \frac { { n }^{ 2 }+n }{ 2 } . The price at which the shop buys the apples is equal to the distance traveled times 5, 5 n 5n . From these two equations we understand that the total revenue therefore will be equal to

5 n ( 100 ( n 2 + n 2 ) ) 5n\left( 100-\left( \frac { { n }^{ 2 }+n }{ 2 } \right) \right) The vertex of this expression is ( 8 , 2560 ) \boxed { \quad (8,2560)\quad } Telling us that after eight miles the revenue will be the highest it can be, so to maximize his revenue he should sell his apples to the eighth shop.

Thx. Nicolas for explaining the solution of my problem :)

Arashdeep Singh - 7 years, 3 months ago

you can also do it without drawing its graph...just diffrentiate the expresion and put it equals to zero then find the value of n it will be 8..

Er Amit Sharma - 7 years, 3 months ago

Log in to reply

I haven't done any calculus before, so I'm just going with that which I know :)

Nicolas Bryenton - 7 years, 3 months ago
Aditya Joshi
Mar 6, 2014

Let us first write the money he makes on the n th n^{\text{th}} town per apple .

He makes $ 5 \$ 5 on the first down, $ 5 + $ 5 = $ 10 \$ 5 + \$ 5 = \$ 10 on the second town, $ 5 + $ 5 + $ 5 = $ 15 \$ 5 + \$ 5 + \$ 5 = \$ 15 on the third town. So that is simply 5 n 5n per apple on the n th n^{\text{th}} town.

Now, how many apples does he have to sell at the n th n^{\text{th}} town? Well, if he sold all the apples on the 3 rd 3^{\text{rd}} town, for example, he would have 100 ( 1 + 2 + 3 ) 100 - \left( 1 + 2 + 3 \right) apples remaning. So on the n th n^{\text{th}} town, he has 100 ( 1 + 2 + + n ) 100 - \left( 1 + 2 + \dots + n \right) apples remaning, which is simply 100 ( n ( n + 1 ) 2 ) 100 - \left( \dfrac{n(n+1)}{2} \right) .

Thus, the net amount he makes by selling apples at the n th n^{\text{th}} town is, as a function of n n ,

f ( n ) = 5 n ( 100 ( n ( n + 1 ) 2 ) ) \Rightarrow f(n) = 5n \left( 100 - \left( \dfrac{n(n+1)}{2} \right) \right)

We need to find the maximum value of f ( n ) f(n) . Thus, by taking the derivative, and finding f ( n ) f'(n) and equating it to 0 0 , we find the maximum of f ( n ) f(n) at n = 1 3 + 601 3 7.83 n = - \dfrac{1}{3} + \dfrac{\sqrt{601}}{3} \approx 7.83 . Thus, he would make a maximum profit on the 8 th 8^{\text{th}} town, and thus, our answer is 8 \boxed{8} .

Solved it the same way.. (y)

Nishant Jain - 7 years, 3 months ago

solve 3x^2+2x=200 using given conditions and u will get x = 7.83 approx.. so it may be 7 r 8 substitute those 7 and 8 u will find 8 will be required shop

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