Hi Lola,

This looks like a minimization problem to me. You can solve this using single variable cal or multiple variable cal.

In single variable calculus:

optimize pi*x^2 + y^2

we know 2*pi*x+ 4*y = 100   =>  y = 25 - pi*x/2

Now we can rewrite optimization function:

pi*x^2 + (25 - pi*x/2)^2

At an optimization point(min or max), first derivative of the function = 0:

d(pi*x^2 + (25 - pi*x/2)^2) / dx = 0 

2*pi*x - pi * (25 - pi*x/2) = 0

 x = 7, y = 14

Take the second derivate:

you will find it is positive, therefore the point is a minimization. There is no global max point. If you wish to visualize this: you can type minimize pi*x^2 + y^2 in 2*pi*x+ 4*y = 100

 https://www.wolframalpha.com/examples/mathematics/applied-mathematics/optimization/

Alternatively you can try the geogrebra 3d graphing app (web or android)

https://www.geogebra.org/3d?lang=en

Thanks; this was very helpful. 

Out of curiosity, what makes you say this is a minimisation problem? Everything in your explanation made sense to me; I just don't quite understand that. But the mathematics problem doesn't say that it's a maximisation problem; I just said that because I was looking for the maximum area (so I could have gotten what to call it wrong). 

We first look for a critical point, i.e. where the first derivative is zero. But this could be a minimum, maximum or neither. To find which one it is, we can check the sign of the first derivative, at points left and right of the critical point. If the sign of the first derivative (i.e. the slope of the function) changes from positive to negative, the curve is concave down (looks like a hump), meaning it is a local maximum. If sign changes from negative to positive, it is concave up -- the point is a minimum. If the sign does not change, it is neither a minimum or a maximum. Alternatively, you can check the sign of the second derivative at the critical point. If it is positive we know that the slope (the first derivative) is increasing. So,  the curve is concave up at the point, i.e. the point is a minimum and vice versa. If the second derivative is zero we cannot decide using the second derivative. 

https://www.khanacademy.org/math/ap-calculus-ab/ab-diff-analytical-applications-new/ab-5-7/v/second-derivative-test

Okay; thanks for the explanation. That makes sense. I think the question asking for the maximum enclosed area through me off a little bit; the explanation with the second derivative test would of course prove whether it's a minimum or maximum point.

Thanks for your prompt replies, in both cases.