C1 Applications of Differentiation

I'm stuck on this question and I'm looking for some help.

The diagram shows a rectangular enclosure, with a wall forming one side. A rope, of length 20 metres, is used to form the remaining three sides. The width of the enclosure is x metres.

i) Show that the enclosed area is, A m^2, is given by A=20x-2x^2.
ii) Use differentiation to find the maximum value of A.

Picture given: http://i.imgur.com/fCO86AI.png

I have no idea what to do, so any help would be useful.

I'm stuck on this question and I'm looking for some help.

The diagram shows a rectangular enclosure, with a wall forming one side. A rope, of length 20 metres, is used to form the remaining three sides. The width of the enclosure is x metres.

i) Show that the enclosed area is, A m^2, is given by A=20x-2x^2.
ii) Use differentiation to find the maximum value of A.

Picture given: http://i.imgur.com/fCO86AI.png

I have no idea what to do, so any help would be useful.

A good start would be working out an expression for the 'long' side, which would then allow you to work out an expression for the area. We know the total length of the rope and we know the lengths of the two 'short' sides so what's the length of the long side?

20-2x?

Yep! So what's the area enclosed by the box?

x(20-2x) = 20x-2x^2

Yep. So you've done the first part, how would you start the second part?

da/dx= 20-2x


and now I don't know what to do

Well firstly you haven't differentiated it correctly and secondly, dA/dx tells you the rate of change of area with respect to X. We know that the maximum area will be reached at the turning point of dA/dx (technically it could be either a maximum or a minimum but this is a quadratic with a positive x^w term so we don't need to worry about that).

oh yeah, silly mistake on my part. It's 4x. Do I move the -4x to the other side and solve for x? so i'll get x=5?

Yes, x=5 when the area is at its maximum. So you just need to sub this value of x into your original expression for the area and you'll have your answer.

Is that it? Thanks for helping me.

Well the answer isn't 5, it's f(5) where f is your area function. What's your answer?

50 metres

Yeah, 50m^2

I'm about 8 years late but i'm doing this question right now. X does not = 5 as, x = 5 when dy/dx = 0. Not when d2y/dx2 = 0. So the value of x to the gradient function was found but not the value of X of the gradient of the gradient function (the maximum value). X is actually equal to 5/2 and the area of the enclosure is 12.5 m^2. Not 50..