Answer:
Maximum dimensions of the field be 300m of length and width be 200 maximum area is [tex]60000 m^{2}[/tex].
Step-by-step explanation:
2L + 3W = 1200
2L = 1200-3W
Dividing both sides by 2
[tex]\mathrm{L}=(600-1.5 \mathrm{W}) \text { we know area of a rectangle is Length } \times \text { Width }[/tex]
Replace L with (600-1.5W) A = W (600-1.5W) A = [tex]-1.5 W^{2}[/tex] + 600W Axis of symmetry, [tex]x=\frac{-b}{2 a}[/tex] in this equation: a=-1.5, b=600 we get W = 200.
[tex]\text { So the length }=600-1.5 \times 200=300[/tex]
[tex]\text { Therefore the maximum area of the field is }-1.5 \times 200^{2}+600 \times 200=60,000 \mathrm{m}^{2}[/tex]
