# I just need these two....

• May 8th 2005, 06:06 PM
I just need these two....
4. A new parking lot is twice the area of the old one. If the length of the old one is 100 ft more than the width and the width is the same as it was, what are the dimensions of the lot?

And now the really hard ones...

1. The area of a rectangular field is 1200 m^2. Two parallel sides are fenced with aluminum at \$20/m. The remaining two sides are fenced with steel at \$10/m. If the cost of the fencing is \$2200, what is the length of each side fenced with aluminum? What is the length of each side fenced with steel?
• May 8th 2005, 06:39 PM
beepnoodle
Length of old lot = width + 100
Area of old lot = width * (width + 100)
Area of new lot = 2 * width * (width + 100)

Since the width of both lots are the same, the 2 must go with the length,
Area of new lot = width *[2(width+100)]
The length of the new lot is 2*(width+100), the width is the same

Now the hard one,
1200 = xy where x is side of aluminum, y is the side with steel

20(2x) + 10(2y) = 2200 note we multiply by 2 since there are two sides each
40x + 20y = 2200
Now, we have a quadratic equation, y = 1200/x
40x + 20(1200/x) = 2200 multiply everything by x
40x^2 + 24000 = 2200x
40x^2 -2200x + 24000 = 0
Now use the quadratic formula and you get x, then plug in to y = 1200/x
• May 8th 2005, 07:27 PM