Calculus of optimization help ):

Question

If I need to sell 400 chairs. The price per chair is 90 dollars up to and including 300 chairs. Above 300, the price will be reduced by 0.25$ (on the whole order) for every additional chair over 300 ordered.

We need to find the largest and smallest revenues I can make under this deal?

I know that the first 300 chairs will generate 27000$ (300x90). I also can generate an equation (Sigma Notation) to represent the second half. I dont know how to make it on here so Ill exlpain it.

100

E (90 - 0.25(K))

K=1

Completed it gives us 7737.5, so in total with all the boxes I get 34737.5 dollars. but how can I find the largest and smallest revenues I can make off this deal?

Would the largest $revenue$ be from 400 Chairs (34737.5) and the lowest technically speaking would be 1 chair (90)?

The full Question Below

You run a small furniture business. You sign a deal with a customer to deliver up to 400 chairs, the exact number to be determined by the customer later. The price will be 90 dollars per chair up to and including 300 chairs, and above 300, the price will be reduced by $0.25 per chair (on the whole order) for every additional chair over 300 ordered. What are the largest and smallest revenues your company can make under this deal? [Be sure to use calculus of optimization to work out your answers and check your second-order condition.

Answer 1

Yes, the function of revenue is

$$R(x)=\begin{cases}90x & \text{if } x\leq300\\x(90-0.25(x-300)) & \text{if }x\geq300\end{cases}.$$

Could you sell $x$ chairs such that $0<R(x)<9$? I mean, when you sell 1 chair you just get $90\$$. But could you sell a big amount of chairs that satisfy the last condition?

Answer 2

Your revenue $R(x)$ if you sell $x$ chairs for $x>300$ is equal to $$\begin{align*}R(x)&=(90-(x-300)\cdot0.25)x\\&=(90-0.25x+75)x\\&=-0.25x^2+165x\\&=-0.25x(x-660)\end{align*}$$ and for $x\le 300$ is equal to $$R(x)=90\cdot x$$ In sum $$R(x)=\begin{cases}90x, & x\le 300\\ -0.25x(x-660), &x>300\end{cases}$$ Now observe that for $x>300$ the revenue is maximized for $x_0=330$ (since the revenue is a quadratic equation with negative coefficient for $x^2$ and with roots $r_1=0$ and $r_2=660$ the max is attained in the midpoint between the roots). Compare this max with the obvious max when $x\le 300$. If there are no restrictions then the minimum is indeed 90 dollars if you sell one chair.