Thread: calculus max/min

Suppose that body temperature 1 hour afer receiving x mg of drug is given by
T(x)=102-1/6 x^2(1-x/9)
for 0<=x<=6.The absolute value of the derivative,|T'(x)| is defined as sensitivty of the body to the drug dosage.Find the dosage that maximize sensitivity

Originally Posted by gracy

Suppose that body temperature 1 hour afer receiving x mg of drug is given by
T(x)=102-1/6 x^2(1-x/9)
for 0<=x<=6.The absolute value of the derivative,|T'(x)| is defined as sensitivty of the body to the drug dosage.Find the dosage that maximize sensitivity

In order to maximize sensitivity, we're interested to see when its derivative is equal to 0. You're given that the sensitivty of the body to the drug dosage is given by T(x)=102-1/6 x^2(1-x/9), which is:

T'(x) = (x^2/18) - (x/3); you are told that it is between 0 and 6, which is obtained by setting this derivative equal to zero.

Now, in order to maximize it, we need to find the derivative of T'(x) and set it equal to 0.

T''(x) = (x/9) - (1/3) = (x-3)/9
0 = (x-3)/9
x = 3

Thus, 3 mg will maximize sensitivity.

Originally Posted by AfterShock

In order to maximize sensitivity, we're interested to see when its derivative is equal to 0. You're given that the sensitivty of the body to the drug dosage is given by T(x)=102-1/6 x^2(1-x/9), which is:

T'(x) = (x^2/18) - (x/3); you are told that it is between 0 and 6, which is obtained by setting this derivative equal to zero.

Now, in order to maximize it, we need to find the derivative of T'(x) and set it equal to 0.

T''(x) = (x/9) - (1/3) = (x-3)/9
0 = (x-3)/9
x = 3

Thus, 3 mg will maximize sensitivity.

I do not agree with such an approach.
Consider the continous function defined as,
on
1)To find the maximum value you need to check the endpoints.

2)To find the maximum value you need to check the interior points where the derivative is zero or is not differenciable.

Note the standard absolute value function,
is not differenciable at, .
Thus, this problem is harder than it seems.