Answer:
The demand function is [tex]\mathbf{D(x) = 1200(\sqrt{25-x^2})+ 124000}[/tex]
Step-by-step explanation:
A firm has the marginal-demand function [tex]D' x = \dfrac{-1200}{\sqrt{25-x^2 } }[/tex].
Find the demand function given that D = 16,000 when x = $4 per unit.
What we are required to do is to find the demand function D(x);
If we integrate D'(x) with respect to x ; we have :
[tex]\int\limits \ D'(x) \, dx = \int\limits{\dfrac{-1200 x}{\sqrt{25-x^2}} } \, dx[/tex]
[tex]D(x) = \int\limits{\dfrac{-1200 x}{\sqrt{25-x^2}} } \, dx[/tex]
Let represent t with [tex]\sqrt{25-x^2}}[/tex]
The differential of t with respect to x is :
[tex]\dfrac{dt}{dx}= \dfrac{1}{2 \sqrt{25-x^2}}}(-2x)[/tex]
[tex]\dfrac{dt}{dx}= \dfrac{-x}{ \sqrt{25-x^2}}}[/tex]
[tex]{dt}= \dfrac{-xdx}{ \sqrt{25-x^2}}}[/tex]
replacing the value of [tex]\dfrac{-xdx}{ \sqrt{25-x^2}}}[/tex] for dt in [tex]D(x) = \int\limits{\dfrac{-1200 x}{\sqrt{25-x^2}} } \, dx[/tex]
So; we can say :
[tex]D(x) = \int\limits{\dfrac{-1200 x}{\sqrt{25-x^2}} } \, dx[/tex]
[tex]D(x) = 1200\int\limits{\dfrac{- x}{\sqrt{25-x^2}} } \, dx[/tex]
[tex]D(x) = 1200\int\limits \ dt[/tex]
[tex]D(x) = 1200t+ C[/tex]
Let's Recall that :
t = [tex]\sqrt{25-x^2}}[/tex]
Now;
[tex]\mathbf{D(x) = 1200(\sqrt{25-x^2}})+ C}[/tex]
GIven that:
D = 16,000 when x = $4 per unit.
i.e
D(4) = 16000
SO;
[tex]D(x) = 1200(\sqrt{25-x^2}})+ C[/tex]
[tex]D(4) = 1200(\sqrt{25-4^2}})+ C[/tex]
[tex]D(4) = 1200(\sqrt{25-16}})+ C[/tex]
[tex]D(4) = 1200(\sqrt{9}})+ C[/tex]
[tex]D(4) = 1200(3}})+ C[/tex]
16000 = 1200 (3) + C
16000 = 3600 + C
16000 - 3600 = C
C = 12400
replacing the value of C = 12400 into [tex]\mathbf{D(x) = 1200(\sqrt{25-x^2}})+ C}[/tex], we have:
[tex]\mathbf{D(x) = 1200(\sqrt{25-x^2})+ 124000}[/tex]
∴ The demand function is [tex]\mathbf{D(x) = 1200(\sqrt{25-x^2})+ 124000}[/tex]
