What is the optimal number of items to produce for maximum profit in business?

Shakattack12
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Homework Statement


A manufacturer makes a batch of n items with the cost (in dollars) of each item being:
n2-6n+35.
The manufacturer sells the items for $50 each. How many items should be produced in each batch to maximise profit.

Homework Equations


Cost = C(n) = n2-6n+35
Revenue = R(n) = 50n

The Attempt at a Solution


Profit = 50n - (n2-6n+35)
= 56n - n2-35
dP/dn = 56 - 2n
let dP/dn = 0
0 = 56 - 2n
n = 28
After testing the nature I found it to be a maximum which is all good but the textbook says the answer is 5. Where did I go wrong?
 
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Shakattack12 said:

Homework Statement


A manufacturer makes a batch of n items with the cost (in dollars) of each item being:
n2-6n+35.
The manufacturer sells the items for $50 each. How many items should be produced in each batch to maximise profit.

Homework Equations


Cost = C(n) = n2-6n+35
Revenue = R(n) = 50n

The Attempt at a Solution


Profit = 50n - (n2-6n+35)
= 56n - n2-35
dP/dn = 56 - 2n
let dP/dn = 0
0 = 56 - 2n
n = 28
After testing the nature I found it to be a maximum which is all good but the textbook says the answer is 5. Where did I go wrong?

Your ##p = n^2 -6n + 35## is the cost per item, so the cost of a batch of ##n## items is ##C(n) = n p##.
 
Thank you so much! Can't believe I missed that.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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