Medication. The number of milligrams x of a medication in the bloodstream t hours after a dose is taken can be modeled by $x ( t ) = \frac { 3000 t } { t ^ { 2 } + 12 }$ $t 0$ . Find the maximum value of x. Round your answer to two decimal places.
A)3.46 mg
B)433.01 mg
C)952.63 mg
D)10.74 mg
E)1125.30 mg

Question

Medication. The number of milligrams x of a medication in the bloodstream t hours after a dose is taken can be modeled by $x ( t ) = \frac { 3000 t } { t ^ { 2 } + 12 }$ $t > 0$ . Find the maximum value of x. Round your answer to two decimal places.
A)3.46 mg
B)433.01 mg
C)952.63 mg
D)10.74 mg
E)1125.30 mg

Quizplus · Accepted Answer

To find the maximum value of $x(t)$, we need to find the derivative of $x(t)$ with respect to $t$ and set it equal to zero to find the critical points. The function given is $x(t) = \frac{3000t}{t^2 + 12}$.The derivative of $x(t)$ with respect to $t$ is $x'(t) = \frac{3000(t^2 + 12) - 3000t(2t)}{(t^2 + 12)^2}$. Simplifying this gives $x'(t) = \frac{3000t^2 + 36000 - 6000t^2}{(t^2 + 12)^2} = \frac{-3000t^2 + 36000}{(t^2 + 12)^2}$.Setting $x'(t) = 0$ to find the critical points gives $-3000t^2 + 36000 = 0$. Solving for $t$ gives $t^2 = 12$, so $t = \sqrt{12} = 2\sqrt{3}$.Substituting $t = 2\sqrt{3}$ back into the original equation to find the maximum value of $x(t)$ gives $x(2\sqrt{3}) = \frac{3000(2\sqrt{3})}{(2\sqrt{3})^2 + 12} = \frac{6000\sqrt{3}}{12 + 12} = \frac{6000\sqrt{3}}{24} = 250\sqrt{3}$.Approximating $250\sqrt{3}$ gives a value close to 433.01 mg, which is option B.

Medication x(t)=3000tt2+12x ( t ) = \frac { 3000 t } { t ^ { 2 } + 12 }x(t)=t2+123000t​

Medication $x ( t ) = \frac { 3000 t } { t ^ { 2 } + 12 }$