Expert: Paul Klarreich - 9/28/2009

Question
QUESTION: A rectangle is bounded by the y-axis and the equation (y/8)^2=-x+7.
Write the area of the rectangle as a function of x, and determine the domain of the function. (Assume the rectangle is ymmetric about the x-axis.) Find the value of x that maximizes the area in the bounded area.


I have seen similar problems but they could not help me with this specific one. Thank you.

ANSWER: Questioner: Michael
Country: United States
Category: Calculus
Private: No
Subject: Rectangle bounded by y-axis
Question: A rectangle is bounded by the y-axis and the equation (y/8)^2=-x+7.
Write the area of the rectangle as a function of x, and determine the domain of the function. (Assume the rectangle is symmetric about the x-axis.) Find the value of x that maximizes the area in the bounded area.
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Your equation is:

(y/8)^2=-x+7

y/8 = sqrt(-x + 7)

y = 8 sqrt(-x + 7)

Now I think your rectangle looks like the picture.  (It might have a lower part, but that would just double it so it does not matter.

Your variables are:

width = x
height = y (or 2y) = 8 sqrt(-x + 7)

Area = xy = 8x sqrt(-x + 7)

And obviously,  0 <= x <= 7

The rest is a routine max-min calculation.


---------- FOLLOW-UP ----------

QUESTION: The Rectangle looks like what picture? The rest is a routine min-max calculation, I'm not sure what that means. How do i find the max? To Write the rectangle as a function of x, does that just mean y = 8 sqrt(-x+7)?

Answer
Sorry about the picture -- I do this all the time; make the picture (it's beautiful) and then forget to attach it.

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As to "a routine max-min calculation"  I assumed, since you didn't say otherwise, that you are doing maximum-minimum problems and knew the general scheme for doing them.

And as to: "To Write the rectangle as a function of x, does that just mean y = 8 sqrt(-x+7)?"

you are not writing the rectangle as a function of x, you are writing the AREA of the rectangle as...

Don't get careless about things like that; this is how students fail calculus.
..........................................

Is this your first attempt at Max-min problems?  If so, the scheme is something like this:

1. Identify the variable(s) in the problem -- the things that can be changed.  Give them names.

2. Write the thing to be Max-ed(or min-ed) in terms of the variable(s).

3. If there are more than one, determine a relationship between the variables (called a 'CONSTRAINT') that will eliminate all but one.  Use a diagram, use your life experience, use your general knowledge and brilliance, do whatever you have to.  This is probably the hardest part.

4. Now differentiate, set the derivative = 0, and solve.  Be sure to check for endpoint max-mins.

AND, Please check the archives for other M-M examples.  There are a lot of them.  Click BROWSE PAST ANSWERS.

So you already have:

width of rectangle = x
height of rectangle = y (or 2y) = 8 sqrt(-x + 7)  << I'll skip the '2'.

Area of rectangle = A = xy = 8x sqrt(-x + 7)  << TO BE MAXIMIZED.

Now go about your business:

dA/dx = 8[ (x)(-1/2 sqrt(-x + 7)) + (sqrt(-x + 7))(1) ]

      - x
= --------------- + sqrt(-x + 7)    << never mind the 8; we don't need it.
 2 sqrt(-x + 7)

Now set
      - x
--------------- + sqrt(-x + 7) = 0
 2 sqrt(-x + 7)

         - x
sqrt(-x + 7) = ---------------
         2 sqrt(-x + 7)

2(-x + 7) = - x
-2x + 14 = -x

14 = x   << that is it.