A company has started selling a new type of smartphone at the price of $110-0.1x where x is the number of smartphones manufactured per day. the parts for each phone cost $50 and labor and overhead for running the plant cost $4000 per day. How many smartphones should the company manufacture and sell per day to maximize profit?

Question

anonymous · Answer

$$
x=\text{number of phones per day}\
\text{selling price per day:}\quad p(x)=110-0.1x\
\text{cost price per day:}\quad c(x) = 50x+4000\
\;\
{\rm profit:}\quad P(x)=p(x)-c(x)
$$

anonymous · Answer

so, what will be the function for profit?

anonymous · Answer

(110-.1x) -(50x+4000)
but i have to take the derivative of it b/c i need to show a sine chart

anonymous · Answer

you have to take the derivative: correct
why: because, the question is asking to find the "x" for which P(x) is maximum

anonymous · Answer

but there is a small mistake I made here.
reading the question again, 
110-0.1x 
id the cost of one phone

anonymous · Answer

so multiply the whole cost by x?

anonymous · Answer

so, the correct selling price:
$$p(x)=x(110-0.1x)=110x-0.1x^2$$

anonymous · Answer

yep

anonymous · Answer

oh... okay thank you so much :)

anonymous · Answer

and you proceed with the first derivative

anonymous · Answer

yes!

anonymous · Answer

what did you get for the critical point?

anonymous · Answer

o'course, in our analysis, we are assuming that all the phones that are manufactured per day will be sold in one day

anonymous · Answer

yes and i got 300 as the critical number

anonymous · Answer

good.
you can now check that P(x) is indeed maximum at x=300 using the second derivative test.

anonymous · Answer

what's the second derivative test?

anonymous · Answer

can't you also check it using a sine chart

anonymous · Answer

you find the second derivative of P(x)
and for each critical point, you check the sign of the second derivative.

what is a sine chart?

anonymous · Answer

its a number line and you plug in the critical number(s) into the function. Then you plug in a number less than your critical number and one greater than the critical number into the function. determine whether the F(x) is greater than or less than that of your F(critical#)

anonymous · Answer

yes. that is the manual version of second derivative.

second derivative tells you the same exact thing without the need to picturize it and in numbers.

anonymous · Answer

oh i like the visual better but that's just my preference

anonymous · Answer

yep. so there you go then.

anonymous · Answer

thanks again!

anonymous · Answer

you are welcome