Drug Concentration The percent of concentration of a drug in...

Drug Concentration The percent of concentration of a drug in the bloodstream $x$ hours after the drug is administered is given by
$$K(x)=\frac{4 x}{3 x^{2}+27}$$
a. Find the time at which the concentration is a maximum.
b. Find the maximum concentration.

Key Concepts

Critical Points

Critical points are values in the domain of a function where its derivative is zero or undefined. These points are candidates for local extrema (maximums or minimums) and are essential in solving optimization problems by determining where the function's behavior changes.

Optimization

Optimization in calculus involves finding the maximum or minimum values of a function within a given domain. This process is key in practical applications, such as determining the time at which a drug reaches its peak concentration in the bloodstream.

Derivative

The derivative represents the rate of change of a function at any point. It allows us to understand whether the function is increasing or decreasing and is fundamental in identifying points where the function may attain a maximum or minimum value.

Quotient Rule

The quotient rule is a technique used to differentiate functions expressed as a ratio of two differentiable functions. It is crucial when working with rational functions, enabling the calculation of the derivative accurately.

Transcript

00:01 We are given the formula for the blood concentration as a function of x, which is hours in time.

00:07 And we want to find the time at which we have a maximum.

00:10 So we'll take the derivative of this function and look for slopes of zero to see when the graph flattens out and has a max or a min.

00:17 And then we could use the quotient rule here.

00:21 So it's going to be low, d high.

00:27 So we'll multiply the bottom times the derivative of the top, which is four, minus.

00:35 High, below, derivative of the bottom is 6x, all divided by the bottom squared.

00:48 I'm going to ignore that because we're only interested in setting the top equal to zero.

00:53 You can ignore the denominator here.

00:55 And then just to jump ahead, so we want to solve this when the slope is zero or the derivative zero, and then just cleaning this up and then factoring, it's going to be the same thing as we're going to get negative 12 x squared because of the positive 12 x squared in the front minus 24 x squared in the back and then plus 24 times 27 times 4 which is 108 and then from here we could go ahead and factor out the 12 or a negative 12 and so that'll end up giving us negative 12 multiply by x squared minus 9 and so finally, from here, we'll end up with two critical points when x is three, positive three or negative three.

01:56 And we're just going to use the positive one since we're only interested in positive times here.

02:02 So we suspect that this will be our max.

02:04 We could prove it really quickly just by plugging some test values into this guy here.

02:11 And so let's look at in between zero and three.

02:15 If we were to plug one in, we'd get a negative times a negative...

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