For some positive constant C, a patient's temperature change, T , due to a dose,D , of a drug is given by T=(c/2-D/3)D^2
What dosage maximizes the temperature change
The sensitivity of the body to the drug is defined as . What dosage maximizes sensitivity?
To find the dosage that maximizes the temperature change, we need to find the maximum of the temperature change equation.
Given the equation for temperature change:
T = (C/2 - D/3)D^2
To find the dosage that maximizes the temperature change, we need to find the derivative of the temperature change equation with respect to the dosage D and set it equal to zero.
Let's find the derivative:
dT/dD = (C/2 - D/3)(2D) + D^2(-1/3)
Simplifying the expression:
dT/dD = CD - D^2/3 - 2CD/6 - D^2/3
Combining like terms:
dT/dD = -2D^2/3 - CD/6 + CD
Setting the derivative equal to zero to find the critical points:
-2D^2/3 - CD/6 + CD = 0
Simplifying the equation:
-2D^2/3 + CD(1/6 - 1) = 0
-2D^2/3 - CD/6 = 0
Factoring out D:
D(-2D/3 - C/6) = 0
Setting each factor equal to zero:
D = 0 (1)
-2D/3 - C/6 = 0 (2)
From equation (1), D = 0 is a critical point, but in this context, we are looking for a positive dosage, so D = 0 is not a valid dosage.
Solving equation (2) for D:
-2D/3 = C/6
D = -3C/12
D = -C/4
However, since D is a dosage and it cannot be negative, we can ignore this solution.
Therefore, there are no critical points, which means there is no maximum or minimum dosage for temperature change. The temperature change will continue to increase as the dosage increases.
Now, let's move on to finding the dosage that maximizes sensitivity.
The sensitivity to the drug can be defined as the absolute value of the derivative of the temperature change equation with respect to dosage.
Let's find the derivative of the temperature change equation again:
dT/dD = (C/2 - D/3)(2D) + D^2(-1/3)
Taking the absolute value of this derivative:
|dT/dD| = |(C/2 - D/3)(2D) + D^2(-1/3)|
Simplifying the expression:
|dT/dD| = |-CD/6 + C/2 -D^2/3|
Since we are looking for the dosage that maximizes sensitivity, we need to find the maximum of the absolute value of the derivative equation.
To find the maximum, we can set the derivative equation equal to zero and solve for D:
-CD/6 + C/2 -D^2/3 = 0
Simplifying the equation:
-C/6D + C/2 -D^2/3 = 0
(3/6)C - (C/6)D -D^2/3 = 0
Multiplying through by 6 to make the equation simpler:
3C - CD -2D^2 = 0
Rearranging the equation:
2D^2 - CD + 3C = 0
This is a quadratic equation in D. We can apply the quadratic formula to solve for D:
D = (-(-C) ± √((-C)^2 - 4(2)(3C))) / (2(2))
D = (C ± √(C^2 + 24C)) / 4
Since the dosage cannot be negative, we ignore the negative square root:
D = (C + √(C^2 + 24C)) / 4
Therefore, the dosage that maximizes sensitivity is D = (C + √(C^2 + 24C)) / 4.
To find the dosage that maximizes the temperature change, we need to find the value of D that maximizes the expression T = (C/2 - D/3)D^2.
To maximize T, we can differentiate it with respect to D and set the derivative equal to zero:
dT/dD = (2C/2 - 3D/3)D^2 + (C/2 - D/3)(2D)
= (C - D)D^2 - (C/2 - D/3)2D
= CD^2 - D^3 - (CD - 2D^2/3)
Setting dT/dD = 0 and solving for D:
CD^2 - D^3 - (CD - 2D^2/3) = 0
D^3 - 2D^2/3 + CD - CD^2 = 0
At this point, finding the exact solution for D can be complex, so we can solve it numerically or approximate using calculus techniques. However, let's simplify the equation and solve it approximately using a technique called the first derivative test:
Let's multiply the equation by 3 to eliminate the fraction:
3D^3 - 2D^2 + 3CD - 3CD^2 = 0
Rearranging the terms:
3D^3 - 3CD^2 - 2D^2 + 3CD = 0
3D^3 - (3C + 2)D^2 + 3CD = 0
Now, let's differentiate this equation with respect to D:
d(3D^3 - (3C + 2)D^2 + 3CD)/dD = 9D^2 - 2(3C + 2)D + 3C
= 9D^2 - 6CD - 4D + 3C
Setting this derivative equal to zero:
9D^2 - 6CD - 4D + 3C = 0
Again, finding the exact solution may be challenging, so let's solve it approximately.
To maximize temperature change, we need to find the value of D that gives a local maximum for T, which means the second derivative of T with respect to D should be negative. Let's differentiate dT/dD one more time:
d^2T/dD^2 = 18D - 6C - 4
Setting this second derivative less than zero:
18D - 6C - 4 < 0
18D < 6C + 4
D < (6C + 4) / 18
D < (3C + 2) / 9
Therefore, the dosage that maximizes the temperature change is D < (3C + 2) / 9.
To find the dosage that maximizes sensitivity, we need to find the value of D that maximizes the expression for sensitivity. However, you have mentioned that the definition of sensitivity is missing from the question. Without the definition, it is not possible to determine the dosage that maximizes sensitivity. Please provide the definition of sensitivity, and I'll be happy to help you further.
dT/dD = (-1/3)D^2 + 2(C/2-D/3)D
= -1/3 D^2 + CD - 2/3 D^2
= CD - D^2
dT/dD = 0 when CD=D^2, or D=C