for a dosage of x cubic centimeters (cc) of a certain drug, the resulting blood pressure b is approximated…

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). [ \begin{align*} B^\prime(x)&=310\times2x-2400\times3x^{2}\ & = 620x - 7200x^{2}\ &=20x(31 - 360x) \end{align*} ]

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)) [ \begin{align*} B(\frac{31}{360})&=310\times(\frac{31}{360})^{2}-2400\times(\frac{31}{360})^{3}\ &=\frac{310\times31^{2}}{360^{2}}-\frac{2400\times31^{3}}{360^{3}}\ &=\frac{310\times961}{129600}-\frac{2400\times29791}{46656000}\ &=\frac{297910}{129600}-\frac{71498400}{46656000}\ &=\frac{297910\times360}{129600\times360}-\frac{71498400}{46656000}\ &=\frac{107247600}{46656000}-\frac{71498400}{46656000}\ &=\frac{107247600 - 71498400}{46656000}\ &=\frac{35749200}{46656000}\ &\approx0.77 \end{align*} ]

Answer:

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