Question 196561
What is formula for finding width and height of rectangle with area of 1250 sq. feet and perimeter of 150 linear feet?

<pre><font size = 4 color = "indigo"><b>
You have to use two formulas:

{{{system(A = LW, P = 2L+2W)}}}

Substitute 1250 for A and 150 for P

{{{system(1250 = LW, 150 = 2L+2W)}}}

We can make the second equation simpler by dividing
both sides of the second equation by 2

{{{system(1250 = LW, 75 = L+W)}}}

Solve the second equation for W by subtracting L
from both sides

{{{system(1250 = LW, 75-L=W)}}}

Substitute {{{(75-L)}}} for {{{W}}} in the
first equation:

{{{1250=L(75-L)}}}

Distribute on the right:

{{{1250=75L-L^2)}}}

Get 0 on the right:

{{{L^2-75L+1250=0}}}

Factor this by writing down all the 
pairs of integer factors of 1250
and the sum of these two factors. Sum
because there is a + before the last
term. (When there is a - before the
last term we us the difference)

factor pairs | sum of factor pairs
---------------------------------
 1 x 1250    |        1251
 2 x 625     |         627
 5 x 250     |         255
10 x 125     |         135
25 x  50     |          75

Now we look at the list of sums to see
if the middle coefficient's absolute
value is among them.  Since the last 
number in the list is 75 and that is
the absolute value of -75, the coefficient
of L, we know the trinomial will factor,
So we rewrite -75L by attaching
the proper signs to 25L and 50L.  For
the sum to come up -75L, we assign the
signs so as to rewrite -75L as
-25L - 50L.

Replace -75L by -25L - 50L

{{{L^2-75L+1250=0}}}

{{{L^2-25L-50L+1250=0}}}

Factor only the first two terms on the left:

{{{L(L-25)-50L+1250=0}}}

Now factor the last two terms on the left:

{{{L(L-25)-50(L-25)=0}}}

Now factor out {{{(L-25)}}}

{{{(L-25)(L-50)=0}}}

Set each factor = 0:

{{{L-25=0}}}, {{{L-50=0}}}

{{{L=25}}}, {{{L=50}}}

Now if the length is 25 feet, then we find the width
by substituting 25 for L in {{{75-L=W)}}}

{{{75-25=50=W}}}

So the width is 50 feet.

Now if the length is 50 feet, then we find the width
by substituting 50 for L in {{{75-L=W)}}}

{{{75-50=25=W}}}

So the width is 25 feet.

So it appears there are two solutions.  One is with the
length of 25 feet and a width of 50 feet.  The other is
with the length of 50 feet and the width of 25 feet.

But of course these are both the same rectangle, and it's
just a matter of which dimension you label "length" and which
you label "width".  So actually there is just one solution.

Edwin</pre>