Word Problem/ Quadratic Function help.?

Question

I need to set up quadratic equations for these three word problems.. (that are solveable by finding the x-intercepts).

please help.. I have a test tomorrow.
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1) The director of a local preschool plans to enclose a rectangular area for a playground. One side will be the side of the building. If 60 ft. of fencing are to be used, what is the maximum area that can be enclosed?

2) What are the dimensions or a rectangle having the greatest area with a perimeter of 20 feet?

(as you can tell, those first two are pretty similar, so if you can do just one, I can probably figure out the other one.)

and 3) What is the largest value that can be obtained by multiplying two real numbers whose sum is three? Round your answer to the nearest hundredth.

Thanks in advance..

hmm okay let me help you help me..

the final answers should look something like:

ax^2 + bx + c = y
(for example: 0.25x^2 - 10x + 800 = y)

Answer 1

1) the fence will make 3 sides so
2L + M =60 so M = 60 -2L
the area
A =LM
so A = L(60 -2L) = 60L - 2 L^2
for A to be maximum we get A'=0
A'=60-4L =0
L=15
so M=60-30=30
so the maximum area =30*15= 450 ft^2

2)
if perimeter=20 so 2(L+M)=20 M=10-L
A=LM =10L-L^2
for maximum area A' =0
A' =10-2L=0
L=5
M=10 - L=5
Maximum area =5*5=25 ft^2

3)
suppose the numbers to be S & R ,, & the number of multplication is N
S+R=3
N= SR
N=S(3-S) =3S-S^2
for N to be the largest put N' =0
N' =3 -2S=0
S=3/2
& R=3-3/2=3/2
N=(3/2)*(3/2) =9/4 =2.25

Answer 2

The maximum area is found in a square.
1) So if one side is the building, divide 60 by 3.
2) For greatest area, divide 20 by 4.

Answer 3

1: Let the sides be l and w
So 2w + l = 60

Area = w*l
Since l = 60 - 2w
Area = w(60-2w) = 60w - 2w^2

2: Let sides be l and w
20 = 2l + 2w
so l = 10 - w

Area = lw = (10-w)w
Area = 10w-w^2

3 a + b = 3 so a = 3-b
Product = ab = (3-b)b = 3b - b^2

Answer 4

sry cant help you, but i do kno the quadratic formula. -b plus or minus the square root of b^2-4ac, all over 2a.
does this help?

Answer 5

to have a rect u need to pairs of equal and opposite so

60x59