calculuseconomicsoptimization
Suppose that it costs a company 5000€ to produce a machine and that the demand for machines (in thousands) for a price of thousand euros is expressed by q(p)=50 − 2p. What price for a machine will maximize the company´s profit?
Please clarify how could I answer these type of questions.If there are any tutorials, I'd be grateful if you share them.
The revenue function of the firm is $R(x)=AR\cdot x=p(x)\cdot x=150x-3x^2$
And the total cost function is $C(x)=TC=0.1x^3 - 3x^2 + 30x$
Therfore the profit function is $P(x)=R(x)-C(x)=150x-3x^2-(0.1x^3-3x^2+30x)=120x-0.1x^3$
To get an extreme point, $x^*$, we have to calculate the derivative of $P(x)$ and set it equal to zero:
$P'(x)=120-0.3x^2=0 \qquad $
Adding $0.3x^2$ on both sides.
$120=0.3x^2$
Dividing both sides by $0.3=\frac3{10}$
$400=x^2$
$x_{1/2}=\pm 20$
Quantities are always positive. Thus $x^*=20$
For the last you have to check if this point is a (relative) maximum.
$P''(x^*)<0 \Rightarrow x^* \texttt{is a maximum}$
You have made a typo at the exercise. $\text{It is "A builder has purchased 21,00}$ $\color{red}{0}$ $\text{square meters of land}$ $\text{on which it is planned to build... "}$
But you have written it at your constraint: $600X+300Y \leq 21,000 \ \ $ (Square meters Constraint)
If you use this constraint the problem is feasible. This is what I got by using the Excel solver:
Excel sheet with formulas
Input mask
Best Answer
This is a simple profit maximization question from Economics. You are given a demand function, here $q(p)=50 − 2p$ and a cost function $5q$. So now you need to put them together to get the profit function.
The profit function in general is given by $$ p(q)q-C(q) \text{ or, less often as } q(p)p-C(p)$$
The profit function should be expressed as a maximization problem in either the price $p$ or quantity $q$. That's for the background.
This should be explained in any intermediate Microeconomics textbook. For example you can check a book by Varian "Intermediate Microeconomics: A Modern Approach"
In your particular case you need to invert first the demand function to express price in terms of quantity. So $$q=50-2p $$ can be inverted to $$ p = 25-\frac{1}{2}q $$
Now, put everythhing together to construct profit function $$ p(q)q-C(q)$$ or $$ (25-\frac{1}{2}q)q - 5q$$
Your problem is now simplified to $$ max_q \left[\left(25-\frac{1}{2}q\right)q - 5q\right] $$
Taking first-order condition you get $$ 25-q-5=0$$ or $$ q = 20 $$
Since the function is concave (second derivative is negative) this is the optimal quantity that maximizes profits.
The price corresponding to q=20 is $p=25-\frac{1}{2}*20=15$.
Note: It is always important to remember whether you solved for optimal q or optimal p. Then use the demand function to obtain the other variable of interest (in the example I solved for q and then used demand function to obtain p).