Applications of Differentiation (Max Min problem)

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Skelly4444

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A sweet manufacturer estimates that if it sets the price of a box of chocolates at £p, it will sell n boxes per year.
Where n = 1000(84+12p-p^2) for 2.5<p<15.

(a) Find the price that maximises the number of boxes sold. (I get this to equal 6)
(b) Write down the revenue by selling n boxes at £p. (I get this to be £np = 1000(84+12p-p^2)

Both the above answers agree with the back of the book and I understand them fully. It is part (c) of this question that has completely baffled me.

(c) Hence show that the price that will maximise the manufacturer's revenue is £10.50 to the nearest 50p. Where does this value come from?

Any advice would be greatly appreciated.
 
A sweet manufacturer estimates that if it sets the price of a box of chocolates at £p, it will sell n boxes per year.
Where n = 1000(84+12p-p^2) for 2.5<p<15.

(a) Find the price that maximises the number of boxes sold. (I get this to equal 6)
(b) Write down the revenue by selling n boxes at £p. (I get this to be £np = 1000(84+12p-p^2)

Both the above answers agree with the back of the book and I understand them fully. It is part (c) of this question that has completely baffled me.

(c) Hence show that the price that will maximise the manufacturer's revenue is £10.50 to the nearest 50p. Where does this value come from?

Any advice would be greatly appreciated.
 
For (b), surely you meant "Revenue = np = 1000p(84+12p-p^2)". You omitted the p in the final formula.

Now for (c), you have to maximize this new function. Distribute the factor p, and differentiate again.
 
I'm sorry but I still don't understand how we can show that the price that maximises the manufacturer's revenue is £10.50 to the nearest 50p.
Where does this value come from? Differentiating this expression doesn't produce a value close to 10.5. Please can you clarify how they've obtained this value.
Thanks.
 
I'm not sure which piece you are missing; it would help if you showed what you got when you differentiated.

Let's start it. We have

Revenue = 1000p(84 + 12p - p^2) = 1000(84p + 12p^2 - p^3)​

What do you get when you differentiate that?

When is that 0?

You won't get exactly 10.50; I get about 10.63, which they then round to the nearest 0.5, getting 10.50.
 
Do we just differentiate the quadratic ignoring the common factor of 1000?
This gives me 84+24p-3p^2.
If we set this equal to zero, I still don't obtain the value of 10.63, what am I doing wrong here?
 
Skelly4444 said:
Do we just differentiate the quadratic ignoring the common factor of 1000?
This gives me 84+24p-3p^2.
If we set this equal to zero, I still don't obtain the value of 10.63, what am I doing wrong here?
Click to expand...
It is not a quadratic!!! But I see that you did differentiate the cubic polynomial, so you aren't as far off as I thought momentarily. You just used the wrong word.

As for "ignoring the 1000", you can, in a sense -- if you understand why. What did you do for part (a)? This isn't really any different.

You can either (a) distribute the 1000 and then differentiate (as I already distributed the p for you); or (b) you can divide the function by 1000 knowing that will not change the location of the extrema; or (c) you can use the fact (which should be natural to you once you have done calculus long enough) that [MATH]\frac{d}{dx}(ku) = k \frac{du}{dx}[/MATH]. The derivative is 1000(84+24p-3p^2), and this is zero when 84+24p-3p^2 is zero.

Now, what DID you get when you solved this equation? In order to tell you what you are doing wrong, I need to see what you are doing. One possibility is that you only looked at one of the two solutions, and it was the wrong one; if you'd shown work, or even just the answer you got, I'd know by now. So please do so now. Then I can probably give you all sorts of good advice.
 
Thanks for all your help. I actually do get the answer of 10.63 now. I must have typed something into the calculator incorrectly initially.
You're quite right, it isn't a quadratic, it's a cubic, not sure why I said it was a quadratic.

I understand what you're saying with regard to the common factor as I've done a bit of calculus before. Thanks once again for all your help.
Regards
Simon
 
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