2.3 Models and applications  (Page 3/9)

This is a distance problem, so we can use the formula $d=rt,$ where distance equals rate multiplied by time. Note that when rate is given in mi/h, time must be expressed in hours. Consistent units of measurement are key to obtaining a correct solution.

First, we identify the known and unknown quantities. Andrew’s morning drive to work takes 30 min, or $\frac{1}{2}$ h at rate $r.$ His drive home takes 40 min, or $\frac{2}{3}$ h, and his speed averages 10 mi/h less than the morning drive. Both trips cover distance $d.$ A table, such as [link] , is often helpful for keeping track of information in these types of problems.

Write two equations, one for each trip.

As both equations equal the same distance, we set them equal to each other and solve for r .

We have solved for the rate of speed to work, 40 mph. Substituting 40 into the rate on the return trip yields 30 mi/h. Now we can answer the question. Substitute the rate back into either equation and solve for d.

The distance between home and work is 20 mi.

The perimeter formula is standard: $P=2L+2W.$ We have two unknown quantities, length and width. However, we can write the length in terms of the width as $L=W+3.$ Substitute the perimeter value and the expression for length into the formula. It is often helpful to make a sketch and label the sides as in [link] .

Now we can solve for the width and then calculate the length.

The dimensions are $L=15$ ft and $W=12$ ft.

The standard formula for area is $A=LW;$ however, we will solve the problem using the perimeter formula. The reason we use the perimeter formula is because we know enough information about the perimeter that the formula will allow us to solve for one of the unknowns. As both perimeter and area use length and width as dimensions, they are often used together to solve a problem such as this one.

We know that the length is 6 in. more than the width, so we can write length as $L=W+6.$ Substitute the value of the perimeter and the expression for length into the perimeter formula and find the length.

Now, we find the area given the dimensions of $L=15$ in. and $W=9$ in.

The area is $135$ in. ² .