How does the equation y = 3 + (4/2)x represent Bob's cake baking rate?

[mathdad]

Bos has 3 cakes and bakes 4 cakes every 2 hours.

The textbook tells me that the correct algebraic interpretation of the above statement is y = 3 + (4/2)x.

I cannot see the connection between the words and this linear equation.

Can someone explain in what way does the equation represent the statement about Bob and his cakes?

The author decided to use y.

I can see that "3 cakes and" = 3 +.

I do not understand how 4/2 represents "4 cakes every 2 hours."

What does x represent here?

What does y represent here?

 

[MarkFL]

I don't like this problem because the number of cakes Bob has at some given time would be a discrete function, not continuous. So, let's use a problem that is mathematically equivalent, but lends itself more readily to a continuous function.

Initially, Bob finds that he is 3 miles from home and moving at a constant velocity he finds that for every 2 hours that goes by, he is 4 miles farther from home...find Bob's distance from home in miles as a function of time in hours. And so we would have:

\(\displaystyle D(t)=\frac{4\text{ mi}}{2\text{ hr}}t\text{ hr}+3\text{ mi}=(2t+3)\text{ mi}\)

Does that make more sense?

 

[mathdad]

I understand parts of your word problem. For example, I know where D(t) comes from. I do not understand the conversion factor of 4 miles/2 hours. I know why (4/2)t became 2t but the words "for every 2 hours that goes by, he is 4 miles farther from home," I do not grasp why this leads to (4/2)t.

 

[MarkFL]

I understand parts of your word problem. For example, I know where D(t) comes from. I do not understand the conversion factor of 4 miles/2 hours. I know why (4/2)t became 2t but the words "for every 2 hours that goes by, he is 4 miles farther from home," I do not grasp why this leads to (4/2)t.

When something changes at a constant rate, the value of that something will be a linear function in terms of the independent variable. For example, if I pull into a gas station to fill my tank, and my tank initially has 3 gallons in it, and the pump is able to deliver fuel at a rate of 4 gallons per minute, then if time t = 0 corresponds to the moment I begin pumping gas, the amount in my tank is:

\(\displaystyle A(t)=4t+3\)

More generally, for different pumps capable of different delivery rates $r$ and for different initial amounts $A_0$, we could write:

\(\displaystyle A(t)=rt+A_0\)

Now, let's arrange this so that we solve for the rate $r$:

\(\displaystyle r=\frac{A(t)-A_0}{t}\)

The numerator represents the amount of fuel delivered in time $t$ (the total amount less the initial amount) and this is divided by $t$ to give the amount that the pump is able to deliver for each or per unit of time...and this is the delivery rate.

Do you now see how the pieces all fit together?

 

[mathdad]

I am slowly able to piece it all together. Word problems are my biggest weakness in math. I feel UNWORTHY of the title MATHEMATICIAN because a math person who cannot solve word problems is just as weak as the most lost pupil.