Question

An apartment complex has 250 departments to rent. If they rent x apartments then their monthly profit, in dollars, is given by $P\left( x \right) = - {\mathbf{8}}{x^2} + {\mathbf{3200}}x - {\mathbf{80000}}$. How many apartments should they rent in order to maximize the profit?

Answer 1

Hint : For any function y(x) the maxima and minima is at the point where the slope of the function y(x) is 0 and it can also be at the end points.
Slope is 0 at the point where derivative of the function y(x) is zero
That is $= \dfrac{{dy}}{{dx}} = y'(x) = 0$
For example if in any function y(x) the x ranges from $a \leqslant x \leqslant b$and the derivative is 0 at point c
Then check the value of y(a), y(b) and y(c).
The one with the maximum value will give you the maxima and with the minimum value will give you minima.

Complete answer : Given,
Given the total no. of department $= 250$
Therefore, the department ranges from [0,250]
That is if x is the no. of departments then x must be in range, $0 \leqslant x \leqslant 250$.
Now as per the question the profit is given by,
$P(x) = - 8{(x)^2} + 3200(x) - 80000$
Differentiating the above quadratic equation, we get,

P\prime (x) = \dfrac{{d( - 8{x^2})}}{{dx}} + \dfrac{{d(3200x)}}{{dx}} - \dfrac{{d(80000)}}{{dx}} \\\ P\prime (x) = - (2) \times 8 \times x + 3200 \\\ P\prime (x) = - 16x + 3200 \\\ \end{gathered} $$ As we know that maxima can be at the point where, P’(x) = 0 We get, $$\begin{gathered} P'(x) = 0 = - 16x + 3200 \\\ \Rightarrow 16x = 3200 \\\ \Rightarrow x = \dfrac{{3200}}{{16}} \\\ \Rightarrow x = 200 \\\ \end{gathered} $$ Now using the above value in P(x) We get, $$\begin{gathered} P(200) = - 8{(200)^2} + 3200(200) - 80000 \\\ \Rightarrow P(200) = - 320000 + 640000 - 80000 = 240000 \\\ \end{gathered} $$ Also, P(0) $$ = - 8{(0)^2} + 3200(0) - 80000 = {\text{ }} - 80000$$ And, P(250) $$ = - 8{(250)^2} + 3200(250) - 80000 = \;220000$$ Hence we can clearly see that P(x) is maximum at $$x = 200$$ Thus, they will get the most profit if they rent 200 apartments out of 250 apartments. **Note** : We can find the maxima and minima by above method only if the function is differentiable for example if y(x)= $$|x|$$then it is not differentiable. Also the slope of any constant value is always 0 that is there value won’t change for any value of x