Homework Help: Calculus

Posted by Anonymous on Tuesday, May 29, 2007 at 9:11pm.

R=M^2(c/2-m/3)

dR/dM=CM-M^2

I found the derivative. Now how would I find the vale of M that maximize the derivative dR/dM?

set it to zero, and solve for m. You get two solutions. Use the second derivative to see which one is the max.

I get M=C. How do I go from there? What do I do?

M=C is one of the solutions for a max/min. Use the second derivative to see which. M=0 is the other.

how can you tell which is the max/min? They are both variables. c=positive constanct, m=amt of medicine absorbed in the blood, dR/dM=sensitivity of the body of medicine

How do you get m=0? from taking the second derivative? does c=1?

Can you check my work on how I get the derivative of R=M^2(c/2-m/3) NEXT R=1/2(CM^2)-1/3(M^3) NEXT dR/dM=CM-M^2. I checked in the back, and that was the answer for the first part of the equation. Isn't the derivative of constant c, zero? I am confused.

Calculus - Amanda, Monday, October 12, 2009 at 9:25am
C is a constant so you use the product rule.

2m(c/2-m/3)+m^2(1/3)

2m being the derivative of m^2. times the original second term. then the derivative of the second term times the original first.

1/3 being the derivative of the (c/2-m/3) because C is a constant

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