My Math Forum Calculus:Chain Rule - Marginal Revenue Product
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 February 8th, 2011, 09:51 PM #1 Newbie   Joined: Feb 2011 Posts: 6 Thanks: 0 Calculus:Chain Rule - Marginal Revenue Product A factory owner who employes m workers finds that they produce q= 1.8m(1.8m+9)^3/2 units of product per day. The total revenue R in dollars is R=1029q / (336672+4q)^1/2 (c) The marginal-revenue product is defined as the rate of change of revenue with respect to the number of employees. Therefore, marginal-revenue product=dR/dm If q and R are given as above then, when m= 15, the marginal-revenue product is _________ ? dollars/(one worker). This means that if employee number 16 is hired, revenue will increase by approximately _________? dollars per day.
 February 8th, 2011, 10:35 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Calculus:Chain Rule - Marginal Revenue Product As this is a calculus problem, I've moved the topic to the Calculus sub-forum. Also, I deleted the duplicate topic, as a question only needs to be posted once. One of the duties of moderating is keeping the forum clear of redundancy. I hope you find this reasonable, as I am not trying to keep you from getting a solution. That being said, welcome to the forum! You are encouraged to post questions, just pick the most appropriate sub-forum and post the question once. To find $\frac{dR}{dm}$ we will need: (1) $\frac{dR}{dq}$ (2) $\frac{dq}{dm}$ (3) $q(15)$ Starting with (3) we have: $q(15)=1.8(15)(1.8(15)+9)^{\frac{3}{2}}=5832$ Continuing with (2) we find: $\frac{dq}{dm}=1.8m$$\frac{3}{2}\(1.8m+9$$^{\frac{1 }{2}}$$1.8$$\)+1.8$$1.8m+9$$^{\frac{3}2{}}=8.1$$1. 8m+9$$^{\frac{1}{2}}$$m+2$$$ When m = 15, then we have: $\frac{dq}{dm}=826.2$ Finally with (1) we find: $\frac{dR}{dq}=\frac{(336672+4q)^{\frac{1}{2}}1029-1029q$$\frac{1}{2}(336672+4q)^{-\frac{1}{2}}\cdot4$$}{336672+4q}=$ $\frac{1029$$336672+2q$$}{$$336672+4q$$^{\frac{3}{2 }}}$ When q = 5832, we have: $\frac{dR}{dq}=1.659434$ Now, using the chain rule, we have: $\frac{dR}{dm}=\frac{dR}{dq}\cdot\frac{dq}{dm}=(1.6 59434)(826.2)\approx1371.02$
 February 17th, 2013, 10:18 PM #3 Newbie   Joined: Feb 2013 Posts: 3 Thanks: 0 Re: Calculus:Chain Rule - Marginal Revenue Product Im a little confused what is the answer to the first blank?

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