Respuesta :
Answer:
a)the weekly profit as a function of price is[tex]P=-10 p^2 + 600 p - 6500[/tex]
b) a bottle of wine be sold at $30 to realise maximum profit
c) the maximum profit that can be made by the producer is $2500
Step-by-step explanation:
The weekly demand curve of a local wine producer is p= 50-0.1q
p = price
q = quantity
Revenue function R =Price \times Quantity
R=(50-0.1q)q
[tex]R=50q-0.1q^2[/tex]
Cost function: c= 1500+ 10q
Profit function=R(x)-C(x)
Profit function= [tex]50q-0.1q^2-1500-10q[/tex]
Profit function= [tex]40q-0.1q^2-1500[/tex] ----1
We have q = 500 - 10 p using p = 50 − 0.1q
[tex]P=-0.1 (500 - 10 p)^2 + 40 (500 - 10 p)- 1500\\P= -10 p^2 + 600 p - 6500[/tex]
General quadratic equation:[tex]ax^2+bx+c=0[/tex]
On comparing
a = -0.1 , b = 40 , c = -1500
Maximum Profit is at q = [tex]\frac{-b}{2a}=\frac{-40}{2(-0.1)}=200[/tex]
To find price must a bottle of wine be sold to realise maximum profit
p= 50-0.1q
p= 50-0.1(200)=30
Substitute the value of q in profit function(1) get the maximum profit
So, Profit function= [tex]40\left(200\right)-0.1\left(200\right)^{2}-1500=2500[/tex]
Hence
a)the weekly profit as a function of price is[tex]P=-10 p^2 + 600 p - 6500[/tex]
b) a bottle of wine be sold at $30 to realise maximum profit
c) the maximum profit that can be made by the producer is $2500
