I was wondering if this the the correct way to answer the question in the image:
$A=(x)=-x^2+1500x=-1(x^2-1500x)=-1x^2-1500x+(750)^2-(750)^2)$ $A=(x)=-(x^2-1500x+(750)^2)+(750)^2=-(x-750)^2+(750)^2$
Which would mean the vertex is $(750,750^2)$ making the maximum area $562,000m^2$
Having shown the answer for part (i), you now have an expression for A.
You have completed the square to obtain an expression which gives you the minimum of your quadratic, and this is fine, and is correct. However I believe from the wording of the question (calculus) that the intended method was to differentiate. This is an answer for part (ii).
$$\frac{dA}{dx}=1500-2x\\\frac{dA}{dx}=0\implies x=750\implies A=1500\cdot750-750^2=2\cdot750\cdot750-750\cdot750=750^2$$ just as you got.
As for part (i), you need to show that the area is $A=1500x-x^2$. You can write the perimeter in terms of $x$ and $y$, and you can also write the area in terms of $x$ and $y$. Simply substituting one equation into the other will give an expression for the area in terms of $x$, and it should match the answer given. This is what the 6 marks are awarded for.
Then what I have done above for part (ii) will yield the other marks.