Question 1173185: A dog trainer has 88 ft of fencing that will be used to create a rectangular work area for dogs. If the trainer wants to enclose an area of 420 ft2, what will be the dimensions of the work area?
Found 4 solutions by math_tutor2020, ankor@dixie-net.com, Alan3354, greenestamps:
Answer by math_tutor2020(3816)   (Show Source):
Answer by ankor@dixie-net.com(22740)  (Show Source):
You can put this solution on YOUR website! A dog trainer has 88 ft of fencing that will be used to create a rectangular work area for dogs.
2L + 2w = 88
simplify, divide by 2
L + w = 44
L = (44-w), use this form for substitution
:
If the trainer wants to enclose an area of 420 ft2, what will be the dimensions of the work area?
L * w = 420
replace L with (44-w)
(44-w)*w = 420
arrange as a quadratic equation
-w^2 + 44w - 420 = 0
multiply by -1
w^2 - 44w + 420 = 0
this will factor to
(w-14)(w-30) = 0
two solutions
w = 14, then 44-14 = 30 is the length
and
w = 30, then 44-30 = 14 is the width
:
Check this:
30 * 14 = 420
and
2(30) + 2(14) = 88
:
Answer by Alan3354(69443)  (Show Source):
You can put this solution on YOUR website! A dog trainer has 88 ft of fencing that will be used to create a rectangular work area for dogs. If the trainer wants to enclose an area of 420 ft2, what will be the dimensions of the work area?
==============================
Area = L*W = 420
P = 2*(L + W) = 88
-------------
L*W = 420
L + W = 44 ---> L = 44-W
-----
W*(44-W) = 420
W^2 - 44W = -420
W^2 - 44W + 484 = -420+484 = 64
(W-22)^2 = 64
W-22 = 8
W = 30
L = 14
Answer by greenestamps(13195)  (Show Source):
You can put this solution on YOUR website!
You have, so far, three responses to your post, showing three different ways to solve the problem algebraically. All three methods are fine; and you should understand how to use all of them.
All three responses start by finding two equations involving the length L and width W:
L + W = 44 (length plus width is half the given perimeter of 88)
L * W = 420 (length times width is the given area of 420)
All three responses then use those equations to find a quadratic equation that can be solved to find the answer to the problem:
From here the solutions use different methods.
One uses completing the square; another uses the quadratic formula. Both of those methods will always work.
The third response solves the equation by factoring, which is probably the most common method for solving the problem. The tutor doesn't show HOW to do the factoring; he simply shows that the equation factors to
You should be able to come up with that factorization yourself.
However, note that this factorization is performed by finding two numbers whose product is 420 and whose sum is 44. But that is EXACTLY what the two equations you started with tell you you need to do -- so the standard algebraic solution method doesn't get you any closer to the solution than you were at the start.
So, if a formal algebraic solution is not required, the simplest and fastest way to finish the problem, starting with the two equations, is simply to find two numbers whose product is 420 that have a sum of 44. (Note this will be faster than finding two numbers whose sum is 44 that have a product of 420, because there are fewer possibilities.)
420 = 42*10; 42+10 = 52 No...
420 = 21*20; 21+20 = 41 No...
420 = 14*30; 14+30 = 44 YES!
|
|
|
