Let x be the width of the fence and y be the length of the fence. Then the perimeter of the fence is x+x+y+y=100. You see that x+y=50 and y=50-x.
The area of the fence is A(x)=x·y=x(50-x),
![A(x)=50x-x^2](https://img.qammunity.org/2019/formulas/mathematics/middle-school/w80i0hdg5yz0e1o7cw8ok0va4ckjcor8x5.png)
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Find the derivative:
![A'(x)=50-2x \\ A'(x)=0 \Rightarrow 50-2x=0, \\ x=25](https://img.qammunity.org/2019/formulas/mathematics/middle-school/tq5rhqxcvx3wottyx32a85sf871z0reptw.png)
For x<25, A'(x)>0 and for x>25, A'(x)<0, this means that x=25 is the maximum point and A(25)=25·(50-25)=25·25=625 is the maximum area.
Answer: maximally he can enclose 625 sq. m area