Find the maximum possible area We have a wire of length 28 meters and we have to form a circle and a square.
What will be the maximum area and how? 
Can we draw the circle?
 A: The circle will have the circumference $C=28 $, meaning that its radius $ r=\frac {C}{2\pi}=\frac {14}{\pi} $, which in turn means that its area $ A_c =\pi r^2=\frac {196}{\pi} $. The perimeter of the square will be $P=28 $, meaning that its side $ a=\frac P4=7 $, which in turn means that its area $ A_s=a^2=49 $. Comparing $ A_c$ and $ A_s $ we get $A_c> A_s$.
A: Let $x$ be the perimeter of the square, and $y$ the perimeter of the circle. So you have $x+y=28$. The area of the square would be $\frac{x^2}{16}$ (why?). The area of the circle would be $\frac{y^2}{4\pi}$ (why?). You did not specify exactly what you want to maximize, but I guess it is the sum of areas:
$$\frac{x^2}{16}+\frac{y^2}{4\pi}.$$
Do you know how to take it from here?
Added: I see no response from the OP, but I'll finish the answer myself in case anyone is interested in this. Plugging $y=28-x$ into the above equation you get
$$\frac{x^2}{16}+\frac{(28-x)^2}{4\pi}=\frac{x^2}{16}+\frac{784-56x+x^2}{4\pi}=
\left(\frac{1}{16}+\frac{1}{4\pi}\right)x^2-\frac{56}{4\pi}x+\frac{784}{4\pi}.$$
Now, you want to find a maximum for the above function for $x\in [0,28]$. Note that since the coefficient of $x^2$ is positive, the function attains its maximum at $x=0$ or $x=28$, so substituting $x=0$ you get $\frac{784}{4\pi}=\frac{196}{\pi}=62.388737\dots$, while substituting $x=28$ you get $\frac{784}{16}+\frac{784}{4\pi}-\frac{784}{2\pi}+\frac{784}{4\pi}=49$.
That is, the answer is that to get the greatest area you should use the whole wire to make a circle.
Note that the answer by @user132181 just computes the area of a circle with perimeter $28$, the area of a square with perimeter $28$ and compares them. He has not shown that this is indeed the maximum sum of areas.  
