Question

Hi! Any one knows how to solve 2C-13-part b? I dont know how to get to the first equation

Answer 1

Hi ehirtz,
I'm not sure what you mean by "the first equation", so let me know if you end up needing more details...
The problem tells us that people will only use a certain amount of kilowatts (x) based on how costly (p) a kilowatt is...
\[ x = 10^5(10-p/2)\]

If the electric company could make electricity and send it to customers for free, then the company could make a lot of money based off the price of a kilowatt times the number of kilowatts their customers use: \[ p*x = p*10^5(10-p/2)\]

Unfortunately for the electric company, the problem tells us it costs them money to create electricity:
\[ cost\ per \ watt= 10-x/10^5\]...here the cost depends on how many watts (x) they end up producing. IE The cost per watt drops as they create more watts (x gets bigger -> cost per watt gets smaller). Since this is the case, we can figure that the total amount the electric company spends each day making electricity will be the total number of watts created times the cost per watt:
\[x* cost\ per \ watt= x*(10-x/10^5)\]

Now we know (1) how much customers pay the electric company, and (2) how much the electric company pays to provide demand to the customs. This gives us a net amount each day...
\[total$$ = p*x-(x* cost\ per \ watt)= p*10^5(10-p/2)-x*(10-x/10^5)\]
...recall that \[ x = 10^5(10-p/2)\] so we substitute this into total$$ and :

Answer 2

\[total$$ = p*10^5(10-p/2)-(10^5(10-p/2))*(10-(10^5(10-p/2))/10^5)\]
\[total$$ =\frac{10^6p}{2}-\frac{10^5p^2}{4}\]

Then solve for where the derivative of total$$ w.r.t. p = 0 (i.e., the critical point)

Answer 3

Thank you! I Got the result. My problem was on that: total$$=p∗x−(x∗cost per watt)=p∗105(10−p/2)−x∗(10−x/105). I think it was an X missing.
Thank you again!