Help with "Largest Area" of Rectangle Problem

[shadow5449]

Can anyone help me with this?

A rectangle has its base on the x-axis and its upper two corner on the graph f(x) = 16 - (x^2). What is the largest area the rectangle can have?

 

[arildno]

What are your own thoughts on the matter?

 

[shadow5449]

Well, I know that the f(x) is going to be an upside down parabola. The area of a rectangle is obviously A = l(w). You would have to find the points on the graph to have your top line...but I'm not sure how you would do this. I'm thinking somthing with the derivative but that's where I'm stuck.

 

[T@P]

try taking a point (x,y) and generaly finding out what the area of the rectangle under will be. then, once you obtain that function, find its maximum. Try looking through your textbook for an example, since all of those kinds of problems are really the same.

 

[arildno]

Do you agree that the area A(x) of the rectangle is given by
$$A(x)=x(16-x^{2}),0\leq{x}\leq{4}$$ ?
How can this help you?

 

[kreil]

A slightly more specific hint:

any point on the graph of a function f(x) is (x,f(x)). Knowing that the corners of the rectangle are on the plot of the given function, what information can this give you about the height of the rectangle? How can this allow you to express A=lw as a single variable function of x?

I hope this helps.

 

[shadow5449]

A = L(h)
A = L(16 - L²)
A = 16L - L³
A` = 16 – 3L²

16 – 3L² = 0
√(L²) = √(16/3)
L = 2.3094011

h = (16 – 2.3094011²)
h = 10.6666

A = 24.63361146

Then I times it by 2 to account for the other side of the parabola (right?)
A = 49.26722297