Question

A firm has the​ marginal-demand function Upper D prime (x )equalsStartFraction negative 1200 x Over StartRoot 25 minus x squared EndRoot EndFraction . Find the demand function given that Dequals16 comma 000 when x equals $ 4 per unit.

Answer 1

Answer:

The demand function is  $\mathbf{D(x) = 1200(\sqrt{25-x^2})+ 124000}$

Step-by-step explanation:

A firm has the​ marginal-demand function $D' x = \dfrac{-1200}{\sqrt{25-x^2 } }$.

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 :

$\int\limits \ D'(x) \, dx = \int\limits{\dfrac{-1200 x}{\sqrt{25-x^2}} } \, dx$

$D(x) = \int\limits{\dfrac{-1200 x}{\sqrt{25-x^2}} } \, dx$

Let represent   t  with $\sqrt{25-x^2}}$

The differential of t with respect to x is :

$\dfrac{dt}{dx}= \dfrac{1}{2 \sqrt{25-x^2}}}(-2x)$

$\dfrac{dt}{dx}= \dfrac{-x}{ \sqrt{25-x^2}}}$

${dt}= \dfrac{-xdx}{ \sqrt{25-x^2}}}$

replacing the value of  $\dfrac{-xdx}{ \sqrt{25-x^2}}}$  for dt in   $D(x) = \int\limits{\dfrac{-1200 x}{\sqrt{25-x^2}} } \, dx$

So; we can say :

$D(x) = \int\limits{\dfrac{-1200 x}{\sqrt{25-x^2}} } \, dx$

$D(x) = 1200\int\limits{\dfrac{- x}{\sqrt{25-x^2}} } \, dx$

$D(x) = 1200\int\limits \ dt$

$D(x) = 1200t+ C$

Let's Recall that :

t  = $\sqrt{25-x^2}}$

Now;

$\mathbf{D(x) = 1200(\sqrt{25-x^2}})+ C}$

GIven that:

D = 16,000 when x = $4 per unit.

i.e

D(4) = 16000

SO;

$D(x) = 1200(\sqrt{25-x^2}})+ C$

$D(4) = 1200(\sqrt{25-4^2}})+ C$

$D(4) = 1200(\sqrt{25-16}})+ C$

$D(4) = 1200(\sqrt{9}})+ C$

$D(4) = 1200(3}})+ C$

16000 = 1200 (3) + C

16000 = 3600 + C

16000 - 3600 = C

C = 12400

replacing the value of C = 12400 into $\mathbf{D(x) = 1200(\sqrt{25-x^2}})+ C}$, we have:

$\mathbf{D(x) = 1200(\sqrt{25-x^2})+ 124000}$

∴ The demand function is  $\mathbf{D(x) = 1200(\sqrt{25-x^2})+ 124000}$