In certain cases in which patients have rapidly
increasing body temperature, powerful drugs are
used to decrease the temperature. Suppose that
t minutes after the injection of such a drug, a
patient’s temperature is y degrees above normal,
where y(t) = −0.006t^2 + 0.36t + 4.2. What is the
patient’s maximum temperature above normal?
this is just a parabola, so its vertex is at t = -b/2a = -.36/-.012 = 30
Since the leading coefficient is negative, the parabola opens downward, so its vertex is the maximum value.
So, what is y(30) ?
To find the patient's maximum temperature above normal, we need to determine the vertex or maximum point of the quadratic function y(t) = -0.006t^2 + 0.36t + 4.2. The maximum point of a quadratic function occurs at the vertex, which has the coordinates (t, y).
The formula for finding the x-coordinate of the vertex of a quadratic function in the form y = ax^2 + bx + c is given by:
t = -b / (2a)
In our case, the coefficients of the quadratic function are a = -0.006, b = 0.36, and c = 4.2. Plugging these values into the formula, we can calculate the x-coordinate of the vertex:
t = -0.36 / (2 * -0.006)
t = 30
So, the patient's maximum temperature above normal occurs 30 minutes after the injection of the drug. Now, let's find the y-coordinate (maximum temperature):
y = -0.006(30)^2 + 0.36(30) + 4.2
y = -0.006(900) + 10.8 + 4.2
y ≈ -5.4 + 10.8 + 4.2
y ≈ 9.6
Therefore, the patient's maximum temperature above normal is approximately 9.6 degrees.