The cost (in dollars) of producing x units of a certain...

The cost (in dollars) of producing $x$ units of a certain commodity is $C(x)=5000+10 x+0.05 x^{2}$ (a) Find the average rate of change of $C$ with respect to $x$ when the production level is changed (i) from $x=100$ to $x=105$ (ii) from $x=100$ to $x=101$ (b) Find the instantaneous rate of change of $C$ with respect to $x$ when $x=100 .$ (This is called the marginal cost. Its significance will be explained in Section $2.7 .)$

The cost (in dollars) of producing $x$ units of a certain commodity is $C(x)=5000+10 x+0.05 x^{2}$
(a) Find the average rate of change of $C$ with respect to $x$ when the production level is changed
(i) from $x=100$ to $x=105$
(ii) from $x=100$ to $x=101$
(b) Find the instantaneous rate of change of $C$ with respect to $x$ when $x=100 .$ (This is called the marginal cost. Its significance will be explained in Section $2.7 .)$

Key Concepts

Finding Maximum and Minimum Rates

In problems involving time-varying rates, analyzing maximum and minimum values is important to understand the behavior of the system. The time when the instantaneous flow rate is greatest or least can provide insights into optimal operating conditions or inherent limits of the system. This concept is integral in applications where knowing the extremes of a process is necessary for design, safety, or efficiency considerations.

Torricelli's Law

Torricelli's Law describes the speed at which a fluid exits an orifice under the force of gravity. It establishes that the velocity of the fluid is proportional to the square root of the height of the fluid above the opening. This principle is often used to relate the height of a liquid in a container to the rate at which it drains, making it fundamental to many problems involving draining tanks or reservoirs.

Instantaneous Rate of Change

The instantaneous rate of change of a quantity, represented as a derivative, gives the rate at which that quantity is changing at any given moment. In the context of a draining tank, it specifically describes the flow rate of water at a particular time. This concept is essential for understanding dynamics in fluid systems and provides insight into how changes occur continuously over time.

Differentiation and the Chain Rule

Differentiation is the process of finding a derivative, which quantifies how a function changes as its input changes. The chain rule is a key technique used in differentiation when dealing with composite functions; it allows one to differentiate a function that is composed of one or more other functions. Together, these tools are crucial for deriving rates of change in models such as those describing the draining of a tank.

Unit Analysis

Unit analysis involves ensuring that all quantities in a calculation are expressed in consistent and appropriate units. When working with fluid flow problems, such as determining the rate at which water leaves a tank, it is important to verify that the units of time, volume, and rate (e.g., gallons per minute) are compatible. This helps confirm that the computed results are physically meaningful and correctly scaled.

Transcript

00:01 With the equation c of x equals 5 ,000 plus 10 x was 0 .05 x squared, and using several different intervals to find the average rate of change of those intervals.

00:11 And we will be using the equation average rate of change equals f of b minus f of a over b minus a.

00:20 So the first interval we will look at is x equals 100 to x equals 105.

00:24 Another way to write this would be 100 comma 105.

00:29 And essentially we will plug both of those numbers into the equation, c of x, and then use that to plug into the equation of the average rate of change.

00:38 So first we'll look at 105.

00:40 The equation will be 5 ,000 plus 10 times 105 plus 0 .05 plus 0 .05 times 105 squared.

00:52 And that will be 6 ,601 .25.

00:57 And for the second number, 100, we will have 5 ,000 plus 10 .5.

01:02 Times 100 plus 0 .05 times 100 squared and that gives us 6500 and to plug into the equation we'll have 6601 .25 minus 6500 over 105 minus 100 and that will give us 101 .25 over 5 which equals 20 .25 and this unit is dollars per unit.

01:33 We don't know what the unit is, so we can just call it unit.

01:37 For the second interval, it will be 100 to 101.

01:41 So the equation for 101 will be 5 ,000 plus 10 times 101, plus 0 .05 times 101 squared.

01:52 And that will be 6 ,500 and 20 .05...

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