Answer:
1857.12 square meters
Step-by-step explanation:
Let L be the length of the rectangle and 'W' be the width
Perimeter = [tex]2L + 2W + 2W[/tex]
The fencing material costs $30 per meter.
The material for the partitions costs $25 per meter
[tex]cost=30(2L + 2W) +25( 2W)\\60L+60W+50W\\60L +110W[/tex]
[tex]7000=60L+110W[/tex]
Solve for L
[tex]7000=60L+110W\\7000-110W= 60L\\L=\frac{700}{6} -\frac{11}{6} W[/tex]
Area = length times width
[tex]( \frac{700}{6} -\frac{11}{6} W)(W)\\A(W)=\frac{700}{6}W -\frac{11}{6} W^2[/tex]
Now take derivative and set it =0
[tex]A(W)=\frac{700}{6}W -\frac{11}{6} W^2\\A'(W)=\frac{700}{6} -\frac{22}{6} W[/tex]
set the derivative =0 and solve for W
[tex]0=\frac{700}{6} -\frac{22}{6} W\\\frac{700}{6}=\frac{22}{6} W\\W= 31.8[/tex]
So width = 31.8 that gives maximum area
[tex]L=\frac{700}{6} -\frac{11}{6} (31.8)=58.4[/tex]
[tex]Area = length \cdot width = 31.8 \cdot 58.4= 1857.12[/tex] square meter
