Applications- Incl. Minimum and Maximum

Question

A patient has an illness that typically lasts about 24 hours. The temperature, T, in degrees Fahrenheit, of the patient t hours after the illness begins is given by: T(t)=−0.03t²+ 0.72t+ 97.7.

Round all answers to 1 decimal place.

When does the patient's temperature reach it maximum value? Answer: After

What is the patient's maximum temperature during the illness? Answer:

Answer 1

In Algebra I and II, it would be inappropriate to discuss derivatives. Derivatives aren't explored until Calculus.

However, the use of graphing to determine minimums and maximums is a common tool. Also, for parabolas, it is known at this level that the min/max occurs when x = -b/(2a), with ax² + bx + c = 0.

 T(t)=−0.03t²+ 0.72t+ 97.7,

So, t = -0.72/(2*(-.03)) = 12 hours after the illness begins.

T(12) = 102.02, which is 102.0, rounded to one decimal place.

No calculus required! And this can easily be checked on a graphing calculator.



Answer 2

If the function is smooth and continuous which this function is then maxima and minima of a function occur when the slope is equal to zero.

This is also defined by when f'(x) = 0

For your function we simply take the derivative with respect to t and set it equal to zero.

T'(t) = -0.06t + 0.72 = 0

t = 12 hours.

This is the only solution to this equation and we can prove to ourselves that this value is a maxima of T(t) rather than a minima or an inflection point by evaluating T''(12).

T''(t) = -0.06

T''(12) = -0.06

Because the second derivative evaluated at 12 is negative then we know the function is concave down at this value so T(12) must be at least a local maxima. We can also determine that this value is the absolute maximum because T''(t) is constant and this function is by definition a parabola.

Now we just need to find T(12) which is 97.7 + 0.72(12) - 0.03(12)^2 = 102.2

So the final answer is the patient has a maximum temperature of 102.2F at 12 hours.