I need help solving this word problem....?

Question

A gardener has 140 feet of fencing to fence in a rectangular shaped vegetable garden. To plant all his vegetables, the gardener needs more than 825 square feet of space. What are all possible values for the length of the rectangle that would meet the space requirement? You may assume that the variable x represents the length of the rectangle.

Answer 1

draw a rectangle. call one side L, and the other W. Notice that 2L + 2W will equal 140, the total available fencing.

Now notice that L+W will equal half that, or 70.

This lets you put things in terms of one variable, L. Because if L+W=70, then W=70-L. So in your drawing you can put 70-L in place of W. Now let's look at the area.

(L)(70-L) needs to be greater than 825. Set up an inequality.

(L)(70-L)>825

Work it out
70L-L^2>825
rearrange terms
-L^2+70L-825>0
multiply by negative one to get rid of the ugly L squared term and remember to reverse inequality.
L^2-70L+825<0
The problem can be done from here by either inspection of factors or quadratic formula. Notice that we want two factors of 825 that sum to -70. In this case that would be (-15) and (-55).

Multiplied together we get +825. Sum them and we have -70. This means the two factors multiplied will be less than 0. (I know it's a little counter intuitive, but that's why algebra is such a powerful intellectual technology.)
(L-15)(L-55) < 0
Separate each factor relative to the product 0. Works by the principle of range. It's a little trickier than when you use the equal sign. (Intuitively? Make it greater than the little number and less than the big one.)
L-15>0 therefore L>15
L-55<0 therefore L<55
Finally we can express it as a value spread:
15L>15. Or L has to be between 15 and 55.

Answer 2

If you're assuming X is the length of the field.
Since it's not a square field, you'll need a variable for the width.... I'll assume Y.

Now, what are the two fact that you know about the field?
140 feet of fencing to use. (The perimeter of the field.)
825 sq. feet of field area must be contained. (The area of the field.)

Do you remember the formula for the area of a rectangle?
( Area = Length * Width )
Is there any formula for the perimeter? (Think about if you had to walk around that field. Walk North, Walk East, Walk South, Walk West... back to your starting point.) Now, how much distance did we say you had to walk??

I hope that helps, without actually doing the work for you.
Good Luck!

Answer 3

If x ft is the length of one side of his garden, then 70 - x ft. is the length of the adjacent side, and the area is x(70 - x) ft^2.

He requires:
x(70 - x) > 825
x^2 - 70x + 825 < 0
(x - 35)^2 - (35^2 - 825) < 0
(x - 35)^2 - 400 < 0
x - 35 < 20
x < 55, and (70 - x) > 15.
The shorter side must be over 15ft, and the longer under 55ft.

Answer 4

15*55 = 825 sq ft

20*50 = 1,000 sq ft

25*45 = 1,125 sq ft

30*40 = 1,200 sq ft

That should help you get started. All have a perimeter of 140 feet.

Answer 5

15 < x < 55 contains all possible value of x

Area = x(70-x) = 70x -x^2
825 < 70x - x^2
15 < x
x < 55