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Homework Help: Optimization/Related Rates ( ) !

  1. Dec 18, 2008 #1
    Optimization/Related Rates (URGENT) !


    Please take a look. Thanks a lot.
  2. jcsd
  3. Dec 18, 2008 #2
    Re: Optimization/Related Rates (URGENT) !

    Average cost is C(x)/x.
  4. Dec 18, 2008 #3
    Re: Optimization/Related Rates (URGENT) !

    yeah. i did do that already ...

    orig equation =25600 +300x+x^2
    avg cost= 25600/x +300 +x

    to find the minimal avg cost , take derivative of avg cost equation and set it to 0.

    ... could you explain what i did wrong instead of just saying average cost is c(x)/x ? i already knew that
  5. Dec 19, 2008 #4
    Re: Optimization/Related Rates (URGENT) !

    As far as your #4 goes you are totally on the right track. You have to look at what you solve for though. You solved for w = 1154.7. What you want is p.
  6. Dec 19, 2008 #5


    Staff: Mentor

    Re: Optimization/Related Rates (URGENT) !

    Hi Ken,
    Here's #1. Let's leave the variables as they were in your formula A = 1/2 * b * h. You converted (counterintuitively) h to x and b to y. I would have done the opposite if I had to change, since y is usually vertical distance and x is usually horizontal distance. It didn't make your work wrong, but it makes it just a bit harder to think about this problem if you have to do some mental translation in addition to the calculus. At any rate, there's not any good reason to switch from b and h to x and y.

    You're given dh/dt = 2.5 cm/sec and dA/dt (not dA/dx) = 4 cm^2/sec. You want to find db/dt at the instant that h = 11.5 cm and A = 99 cm^2. Before you substitute these values in, you need to solve for db/dt.

    A = 1/2 * b * h
    ==> dA/dt = 1/2 (b*dh/dt + h * db/dt)

    Solve the last equation for db/dt, and then substitute the values for h, dh/dt, A, and dA/dt.

    For #2, you want the rate of change of A with respect to s, not with respect to time. IOW, you want dA/ds.

    For #3, as mutton said, the average cost is C(x)/x. Find the equation of C(x)/x, and then take its derivative (with respect to x). Set this to zero and solve for x, the production level. For part e of this problem, evaluate C(x) for the production level you found in part d.

    For #4, this is a calculus problem. You have p = 2l + 3w, and l = 2000000/w. Substitute l in the second equation into your p equation to get p as a function of w. Now find p'(w). Presumably (you should check) this will give you the minimum amount of fencing.
  7. Dec 19, 2008 #6
    Re: Optimization/Related Rates (URGENT) !

    thanks i got all the answers except for 1)

    so heres what i did:

    dh/dt = 2.5 cm
    dA/dt= 4cm
    A= 99

    and i need to find the rate of change of db/dt when h= 11.5

    first i found out what the base would be so I plugged h=11.5 into A= 1/2bh and got b=17.2173913

    Next I took dA/dt 1/2 (b*db/dt + h *dh/dt)

    Is this correct? Next I plugged in all the known variables and isolated db/dt

    db/dt= -1.2051767 ? but it is wrong ...

    thanks again
  8. Dec 22, 2008 #7


    Staff: Mentor

    Re: Optimization/Related Rates (URGENT) !

    So far, so good.
    No, the part in parentheses is wrong. Assuming that both b and h are differentiable functions of t, what do you get for d/dt(b*h) using the product rule?
    At the moment in time of interest, I get db/dt [tex]\approx[/tex] -3.047 cm/min, meaning that the base is decreasing in length.
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