Path around rectangular fountain A landscaper is designing a rectangular fountain with a 4-foot-wide path around it. The equation $A = 4p + 64$ will relate the area $A$, in square feet, of the path to the perimeter $p$, in feet, of the fountain. In the design, how many feet will the perimeter of the fountain increase for each additional square foot of the path’s area?
The answer is 1/4, but I do not know how. Anyone please show me with every vital steps?
Thank you!
 A: It is the inverse of the coefficient $4$ of $p$.  If you add $1$ to $A$, you have to add $\frac 14$ to $p$ to maintain the equality.
A: This question, like many SAT questions, is just asking you to think about the meaning of the slope of a line. 
In this case, the problem is asking: if you increase $A$ by 1, by how much does $p$ increase? Part of the test is whether you recognize that this is precisely the kind of question the slope of a line helps us answer.
You are given $A$ as a linear function of $p$, namely $A=4p+64$, so you can easily read off the slope, which is 4. But what does this mean? Saying the slope is 4 means that when $p$ increases by 1, $A$ increases by 4. In other words, the ratio
$$\frac{\text{change in A}}{\text{change in p}}$$
is 4 to 1.
Hence if we want $A$ to increase only by 1, we should increase $p$ only by $1/4$.
A: You have
$A = 4p+64$.
If the area increases by $1$,
and the new perimeter is
$q$,
then
$A+1 = 4q+64$.
Subtracting these,
$1
= (A+1)-A
=(4q+64)-(4p+64)
=4q-4p
=4(q-p)
$.
Therefore,
$q-p$,
which is the amount that
the perimeter increased by,
satisfies
$q-p = \dfrac14
$.
This is where the answer comes from.
A: There are correct and good answers here.  Here's a simpler one that's also a test taking strategy.  (Some regulars on this site might not appreciate this approach but I've frequently seen this strategy [successfully] taught.)
You want to know the following:

If I increase $A$ by $1$, how much will $p$ increase?

We could analyze it in general.  But because there's only one right answer (which we can infer from the question's phrasing, and it's especially clear if it's a multiple choice question), we can just plug in specific and convenient values and see what happens!
What makes a value convenient?  It really depends on the problem.  In this case we would like to pick a value of $A$ such that solving $A = 4p+64$ for $p$ is "easy."  In theory, any value of $A$ will work, but because $4p$ and $64$ are both multiplies of $4$, then a convenient value of $A$ will be any multiple of $4$ large enough to give us a positive $p$.
So let's try $A=80$.  This gives us $$80 = 4p+64.$$ Solve this to get $p=4$.  Now let's add $1$ to $A$ and see what happens.
If we have $A=81$, then we get $$81=4p+64.$$ So then $4p=17$, and so $$p =\frac{17}4 = 4.25 = 4+\frac14.$$
Therefore, $p$ increases by $1/4$ whenever$A$ increases by $1$. 
A: The area of the path is given by $A = 4p+4\times (4\times 4)$: $4p$ makes for the four segments of path along each side of the rectangle, and you add four times a $4\times 4$ "square feet" squares to connect the segments at corners.
If you enlarge the base rectangle, you will change the perimeter, hence the surface of the four segments. But the four  $4\times 4$ "square feet" squares won't change. So the surface of the path will vary only from the $4p$ term. In other words, the difference between the area for two designs varies $4$ times faster than the perimeter. This linear dependence tells you that if the perimeter varies by $1$, the area will vary by $4$, whatever the initial area. This is a proportional effect, or the slope some people here are asking for. So for the area to vary by 1, the perimeter should vary by $1/4$. We are now ready for equations.
If the surface increases by 1 square foot, i.e. $A' = A+1$, what will be the new perimeter $p'$? $A' =4p'+64$, and you know that $A =4p+64$, so $4p+64+1 = 4p'+64$, which reduces to $4(p'-p)=1$, so the perimeter variation is $p'-p =1/4$.
