Need Help? Read It Submit Answer 11. [0/9 Points] DETAILS MY...

Need Help? Read It Submit Answer 11. [0/9 Points] DETAILS MY NOTES LARAPCALC10 4.6.045. PREVIOUS ANSWERS ASK Revenue A small business assumes that the demand function for one of its new products can be modeled by $p = Ce^{kx}$. When $p = \$50$, $x = 800$ units, and when $p = \$45$, $x = 1300$ units. (a) Solve for C and k. (Round C to four decimal places and k to seven decimal places.) C = 151.85 k = (b) Find the values of x and p that will maximize the revenue for this product. (Round x to the nearest integer and p to two decimal places.) x = 719 units p = $

Transcript

00:01 In this question, we're told that the demand function is c times e to the k x.

00:07 And we are given that p of 800 is 45 and p of 1 ,500 is 40.

00:12 And we are asked to find the constants c and k.

00:17 So from the first condition, p of 800 equals to 45.

00:23 On the other hand, if we plug in x equals 800 in the equation for p, we're going to get c times e to the x.

00:31 Now if we plug in x equals 1500, we are going to get c times e to the 1500k and this gives us a system of equations for finding c and k.

00:48 C times e to the 800k equals to 45 and c times e to the 1500k equals to 40.

01:03 What we are going to do now is solve the equation for the first equation for c to get c equals to 45 divided by 8 e to the 800k.

01:19 And plug in in the second equation.

01:23 In the second equation, we are going to get 45 divided by e to the 800k, multiplied by e to the 1 ,500k equals to 40.

01:38 Then e to the 1500k over e to the 800k simplifies to e to the 1500k minus 800k then we can divide both sides by 45 and simplify the exponent to get e to the 700k equals to 40 over 45 and that simplifies to 8 over 9 now we will rewrite the in equivalent form using logarithms, then by the properties of logarithms, a l 'n of e to the 700k equals to 700k, and from that equation, k equals to a length of 8 over 9 divided by 700, and that's approximately, that's approximately negative 0 .30 -160's and 60 negative 0 .3 zeros and 1683.

03:22 So we found the value of k and now we can use it to find the value of c from this formula.

03:31 C equals to 45 divided by e to the 800k.

03:44 And here we will use the exact value of k, ln of 8 over 9 divided by 700.

03:55 Then we can cancel two zeros.

03:59 So we are going to get a c equals to 45 over e to the 8 ln 8 over 7.

04:10 8 over 7, ln of 8 over 9.

04:17 This is approximately e to the 8 over 7 multiplied by ln of 8 over 9.

04:45 All right, so that's going to be approximately 51 .4840.

05:02 51 .484.

05:14 So this is approximate value of the constant c, and therefore the demand equation is p equals to c multiplied by e to the power, kx.

05:33 And we just found that k equals to negative 0 .30s 16 .83x...

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