SOLUTION: Among rectangles with a perimeter of 350 cm, what are the dimensions of the one having maximum area? Lesson is about the application of quadratic equations, and it requires a compl

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Question 552093: Among rectangles with a perimeter of 350 cm, what are the dimensions of the one having maximum area? Lesson is about the application of quadratic equations, and it requires a complete solution. Thank you! :)
Found 3 solutions by neatmath, Theo, richard1234:
Answer by neatmath(302)   (Show Source): You can put this solution on YOUR website!

I know the answer for this before I begin, but let's go through the steps!

We are given a perimeter of 350 cm.

Let L stand for length, and W stand for width. Then,

given

divided both sides by 2

subtracted W from both sides, solved in terms of L

Now, we can use this information to "maximize" our area:







We can now set up a table and pick some values for W to solve this problem.

Of course, W must be a value between 0 and 350 to satisfy our given info:
W=0, then A=0
W=1, then A=174
W=10, then A=1650
W=80, then A=7600
W=90, then A=7650
W=100, then A=7500
W=150, then A=3750

Interesting, no?

We can see that our Area is being maximized somewhere around 90.

And in fact, if you do the math, the maximum area will occur where W=87.5

If W=87.5, then A=7656.25

If W=87.5, then L=87.5 as well.

And you know what? If the length and width of a rectangle are equal, then you have a square by definition.

Given a set perimeter, the maximum area of a rectangle will ALWAYS, ALWAYS occur when the rectangle is in fact a square.

That's how I knew the answer even before we started.

I just needed to divide 350 by 4 (for each of the sides).

And 350/4 is indeed equal to 87.5.

Final answer:

The dimensions of a rectangle with maximum area and a perimeter of 350 cm are length of 87.5 cm and width of 87.5 cm.

This can also be easily done using Calculus, where we go through the same steps that we did above, but then use something called a derivative to more easily calculate the maximum area of the rectangle.

But for now, this is the best algebraic way to solve this type of problem, in my opinion.

*******************************************************

I hope this helps! :)

Email Scott: neatmath@yahoo.com for help with specific problems,

or to inquire about mathematics tutoring via email or other methods.

Paypal is always accepted for single problems or for more intensive tutoring!

Single problems would range from 50 cents to 5 dollars each, depending on their complexity.


Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
a standard form of a quadratic equation has a form of:
y = ax^2 + bx + c
a is the coefficient of the x^2 term.
b is the coefficient of the x term.
c is the constant term.
the graph of the quadratic equation looks like an umbrella.
it's called a parabola.
if a is positive, the parabola points down and opens up.
if a is negative, the parabola points up and opens down.
the first graph below is the parabola pointing down because the coefficient of the x^2 term is positive.
the second graph below is the parabola pointing up because the coefficient of the x^2 term is negative.
the equations used are:
pointing down / opening up:
y = 2x^2 + 7x - 2

pointing up / opening down:
y = -2x^2 + 7x - 2

a parabola has one maximum point or one minimum point.
if the parabola points down then it has a minimum point.
if the parabola points up then it has a maximum point.
how does this apply to your problem?
see below:
the perimeter of your rectangle is equal to 350 cm.
if we let L = length of the rectangle and W = width of the rectangle, then:
perimeter equals 2L + 2W
area equals L * W.
we know that the perimeter 350 cm.
this means that 2L + 2W = 350
in this equation, we can solve for W to get:
W = (350-2L)/2 which can be simplified to:
W = 275 - L
if we substitute for W in the equation for the perimeter of the rectangle, then we get:
2L + 2*(175-L) = 350
if we let x = L, then this equation becomes:
2x + 2*(175-x) = 350.
since we know that the area of the rectangle equals L * W, we can substitute for W in this equation as well to get:
L * (175-L) = Area of Rectangle.
if we substitute x for L, then we get:
x * (175-x) = Area of Rectangle.
if we let y = area of rectangle, then our equation becomes:
y = x * (175-x)
we simplify this equation to get:
y = 175*x - x^2
this is a quadratic equation.
we can graph this equation to get the diagram below:

you can see that this graph will peak at somewhere around 7500.
we can solve for the max/min point of the equation by using the formula of:
x = -b/2a
our equation is:
y = 175*x - x^2 which we can re-write as:
y = -x^2 + 175*x
this is the standard form of the quadratic equation where:
a = coefficient of the x^2 term is equal to -1.
b = coefficient of x term is equal to 175.
our formula for the max/min point is:
x = -b/2a which comes out to be:
x = -175 / (2*(-1)) which comes out to be:
x = 87.5
when x = 87.5, the value of y becomes:
y = -(87.5)^2 + 175*(87.5) which comes out to be:
y = -7656.25 + 15312.5 which comes out to be:
y = 7656.25
since y represents the area of the rectangle, then the maximum area of the rectangle is equal to 7656.25 square cm.
you can create a table of values to see that this is accurate and that the perimeter will always be 350 cm.
this table will look as follows:
length       width       perimeter        area
x            175-x       2x + 2(175-x)    x*(175-x)

45           130         350              5850
85           90          350              7650
87.5         87.5        350              7656.25 ***** (max)
90           85          350              7650
125          50          350              6250
165          10          350              1650

this quadratic equation had a maximum value because the coefficient of the x^2 term was negative.



Answer by richard1234(7193)   (Show Source): You can put this solution on YOUR website!
Recall that the max/min of a quadratic f(x) = ax^2 + bx = c occurs at x = -b/2a.

Dimensions can be assumed to be L and 175-L, since the perimeter is fixed at 350. We want to maximize -L^2 + 175L, in which the maximum occurs at L = 175/2 = 87.5. This makes the rectangle a square.
-------------
Another solution that takes a far different approach. By AM-GM inequality,



This essentially says that the area is at most some constant. Equality occurs if and only if L = 175-L, or L = 87.5 cm.

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