Sales of a product, under relatively stable market...

Sales of a product, under relatively stable market conditions but in the absence of promotional activities such as advertising, tend to decline at a constant yearly rate. This rate of sales decline varies considerably from product to product, but it seems to remain the same for any particular product. The sales decline can be expressed by the function $$ S(t)=S_{0} e^{-a t} $$ where $S(t)$ is the rate of sales at time $t$ measured in years, $S_{0}$ is the rate of sales at time $t=0,$ and $a$ is the sales decay constant. (a) Suppose the sales decay constant for a particular product is $a=0.10 .$ Let $S_{0}=50,000$ and find $S(1)$ and $S(3)$ to the nearest thousand. (b) Find $S(2)$ and $S(10)$ to the nearest thousand if $S_{0}=80,000$ and $a=0.05$

Sales of a product, under relatively stable market conditions but in the absence of promotional activities such as advertising, tend to decline at a constant yearly rate. This rate of sales decline varies considerably from product to product, but it seems to remain the same for any particular product. The sales decline can be expressed by the function
$$
S(t)=S_{0} e^{-a t}
$$
where $S(t)$ is the rate of sales at time $t$ measured in years, $S_{0}$ is the rate of sales at time $t=0,$ and $a$ is the sales decay constant.
(a) Suppose the sales decay constant for a particular product is $a=0.10 .$ Let $S_{0}=50,000$ and find $S(1)$ and $S(3)$ to the nearest thousand.
(b) Find $S(2)$ and $S(10)$ to the nearest thousand if $S_{0}=80,000$ and $a=0.05$

Key Concepts

Initial Value

The initial value, represented by S?, is the quantity at time t = 0 before the decay process begins. It serves as the starting point for the model, and all subsequent values of the decaying quantity are calculated relative to this initial amount. The initial value is essential in setting the scale of the model and in making accurate predictions about the future state of the system.

Numerical Evaluation of Exponential Functions

Evaluating an exponential decay function at specific time points involves substituting the time variable into the function S(t) = S?e^(–at) and computing the resulting value. This process may require the use of approximations or rounding to obtain practical, real-world estimates, especially when dealing with large or small numbers where exact values are less meaningful.

Exponential Decay Function

Exponential decay describes a process in which a quantity decreases at a rate proportional to its current value. This behavior is modeled by functions of the form S(t) = S?e^(–at), where the decrease is continuous and the rate of change is constant relative to the current amount. This concept is widely applicable in various fields, including physics, biology, economics, and finance, where it helps predict how quantities diminish over time.

Decay Constant

The decay constant, often denoted by 'a', is a key parameter in exponential decay models that determines the speed at which the quantity decreases over time. A higher decay constant means the quantity declines more rapidly, while a lower constant suggests a slower decline. This constant is intrinsic to the process it describes, ensuring the rate of decrease remains proportional to the current value at every moment.

Transcript

00:01 Hello guys, so we're going to answer the question above, which has two parts two.

00:05 The first part is to find s1 and s3 to the nearest thousand and the second part to find s2 and s10 to the nearest thousand.

00:15 So given this formula, let me change the color, blue, okay, for the first part, we have, they gave us a formula of s of s of t because s not e to the negative a t e to negative at so we need to let s of s not here equals 50 000 and a equals 0 .10 to find the first part so to do so we plug in we plug in the values s of 1 s of 1 over here equals 50 thousand three e to the negative a which is write that down a equals 0 .10 negative 0 .10 times one because that that is our that's our t so plugging that in the calculator you get a result of 45 ,241 .871 so that is our s of 1.

01:57 Now we need to find s of 3 so we do the same thing to find s of 3 s of 3 equals 50 ,000 each the negative 0 .10 which is the same a value given to us times 3 you plug this in the in the calculator we get 37 ,000 40 .911 and this is our s3 values this this is it for port a moving on to part b we asked to find the same thing but for two different values first of we have you actually find it for s of 2 and s of 10 and we have a different um we have a different s of o value so let me write that here our a is 0 .05 and our s of o value is 80 000 this time instead of 50 ,000 over here.

03:23 So to do that, to find s of 2, we do s of 2 equals 80 ,000 e to the negative 0 .1 because we multiply, let me just take a step back so you guys can actually understand.

03:44 So it will be negative 0 .05 times 2.

03:53 So, looking that in the calculator, negative 0 .05 times 2.

04:03 We get a value of 72 ,386.

04:14 993...

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