The dosage at which the resulting blood pressure is maximized is x at 0.2 and the maximum dosage is 0.0008.
In the question,
The function is [tex]B(x) = 0. 06x^2 - 0. 2x^3[/tex]
To find the maximum or minimum, take the derivative and set it equal to zero.
⇒ [tex]B'(x) = 2(0. 06)x - 3(0. 2)x^{2}[/tex]
Setting it equal to zero, we get
⇒ [tex]0.12x - 0.6x^{2}=0[/tex]
⇒ 0.6x (0.2-x) = 0
⇒ x = 0 or x = 0.2
Now substitute x = 0.2 in B(x), we get
⇒ [tex]B(0.2) = 0. 06(0.2)^{2} - 0. 2(0.2)^{3}[/tex]
⇒ B(0.2) = 0.0024 - 0.0016
⇒ B(0.2) = 0.0008
To know B(0.2) is maximum, let us find the values for x = 1 and x = 0.01.
For x = 1,
⇒ [tex]B(1) = 0. 06(1)^{2} - 0. 2(1)^{3}[/tex]
⇒ B(1) = 0.06-0.2
⇒ B(1) = -1.04
For x = 0.01,
⇒ [tex]B(0.01) = 0. 06(0.01)^{2} - 0. 2(0.01)^{3}[/tex]
⇒ B(0.01) = 0.000006 - 0.0000002
⇒ B(0.01) = 0.0000058
Thus, x = 0.2 is the maximum.
Hence we can conclude that the dosage at which the resulting blood pressure is maximized is x at 0.2 and the maximum dosage is 0.0008.
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