The area of the field is 9000m^2. Find the dimensions of the field.
2007-01-04 04:32:43 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Okay, letting x,y be sides then 2x+2y=400, and xy=9000.
So, x+y=200, and x=9000/y.
Substitute, and you have (9000/y)+y=200.
Multiply by y on both sides, and you have 9000+y^2=200y.
Or y^2-200y+9000=0.
Use quadratic formula, and you have y=(200+-sqr(200^2-(4*1*9000))/ (2*1) = (200+-63.24555)/2 = 131.623 and 68.3772.
SO, dimensions are 131.623 and 68.3772 approximately.
2007-01-04 05:22:49 · answer #1 · answered by yljacktt 5 · 0⤊ 0⤋
2(l+w)=400
-->l=200-w
l.w=9000
200w-w^2=9000(substitute)
w^2-200w+9000=0
dimensions:131.6x68.4
2007-01-04 13:34:07 · answer #2 · answered by Maths Rocks 4 · 0⤊ 0⤋