Question

for a dosage of x cubic centimeters (cc) of a certain drug, the resulting blood pressure b is approximated by the function below. find the maximum blood pressure and the dosage at which it occurs.
$b(x)=310x^{2}-2400x^{3},0leq xleq0.12$
the maximum is obtained for a dosage of
(round to two decimal places as needed.)

Explanation:

Step1: Find the derivative of \(B(x)\)

Using the power rule \((x^n)^\prime = nx^{n - 1}\), for \(B(x)=310x^{2}-2400x^{3}\), the derivative \(B^\prime(x)=(310x^{2}-2400x^{3})^\prime\).

$$
LATEXBLOCK0
$$

Step2: Find the critical points

Set \(B^\prime(x)=0\). Then \(20x(31 - 360x)=0\).
We have two cases:

• Case 1: \(20x = 0\Rightarrow x = 0\)
• Case 2: \(31-360x=0\Rightarrow x=\frac{31}{360}\approx0.09\)

Step3: Use the second - derivative test

Find the second - derivative \(B^{\prime\prime}(x)\).
Since \(B^\prime(x)=620x - 7200x^{2}\), then \(B^{\prime\prime}(x)=(620x - 7200x^{2})^\prime=620-14400x\)

• When \(x = 0\), \(B^{\prime\prime}(0)=620>0\), so \(x = 0\) is a local minimum.
• When \(x=\frac{31}{360}\), \(B^{\prime\prime}(\frac{31}{360})=620-14400\times\frac{31}{360}=620 - 1240=- 620<0\), so \(x=\frac{31}{360}\approx0.09\) is a local maximum.

Step4: Find the maximum blood pressure

Substitute \(x = \frac{31}{360}\) into \(B(x)\)

$$
LATEXBLOCK1
$$

Answer:

The maximum is obtained for a dosage of \(0.09\) cc.