Question

The rate of growth of profit​ (in millions) from an invention is approximated by Upper P prime (x )equals x e Superscript negative x squared Baseline comma where x represents time in years. The total profit in year 2 that the invention is in operation is $ 25 comma 000. Find the total profit function Upper P (x ). Round to the nearest thousandth if necessary.

Answer 1

Answer:

The function is  $P(x) = \frac{1}{2} e^{-x^2} +0.016$

Step-by-step explanation:

From the question we are told that

    The rate of growth  is  $P'(x) = xe^{x} - x^2$

     The total profit is $P(2) =$$25,000

      The time taken to make the profit is $x = 2 \ years$

         

From the question

     $P'(x) = xe^{-x^2}$  is the rate of growth

  Now here x represent the time taken

Now the total profit is mathematically represented as

       $P(x) = \int\limits {P'(x)} \, = \int\limits {xe^{-x^2}} \,$

So using substitution method

   We have that

                      $u = - x^2$

                      $du = 2xdx$

  So  

        $p(x) = \int\limits {\frac{1}{2} e^{-u}} \, du$

       $p(x) = {\frac{1}{2} [ e^{-u}} +c ]$

       $p(x) = {\frac{1}{2} e^{-x^2}} + \frac{1}{2} c$      recall  $u = - x^2$   and  let  $\frac{1}{2} c = Z$

       

At  x =  2 years  

     $P(x) =$$25,000

So

       Since the profit rate is in million

    $P(x) =$$25,000 = $\frac{25000}{1000000} =$$0.025 millon dollars

So  

       $0.025 = {\frac{1}{2} e^{-2^2}} + Z$  

=>    $Z = 0.025 - {\frac{1}{2} e^{-2^2}}$  

       $Z = 0.016$  

So the profit function becomes

         $P(x) = \frac{1}{2} e^{-x^2} +0.016$