Checking Answer: Maximize area of square under parabola

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SeekerOfDragons

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I'm trying to verify my answer to the following:

Find the area of the largest rectangle with lower base on the x-axis and upper vertices on the parabola y = 20 - 4X^2

Area of Rectangle = Length * Width

Set Length = X
Set Width = Y

A(X) = X * Y
A(X) = X(20 - 4X^2)
A(X) = 20X - 4X^3

A'(X) = 20 - 12X^2

20 = 12X^2
X^2 = 20/12 = 5/3
X = -Sqrt (5/3), Sqrt (5/3)

Using First Derivative Test, determined Sqrt(5/3) is a max.

A(X) = X(20 - 4X^2)
A(Sqrt(5/3)) = Sqrt(5/3)(20 - 4(Sqrt(5/3)^2)
A(Sqrt(5/3)) = Sqrt(5/3)(20 - 4 (5/3))
A(Sqrt(5/3)) = Sqrt(5/3)(20 - 20/3)
A(Sqrt(5/3)) = Sqrt(5/3)(40/3)

Area of Rectangle = 40 Sqrt(15) / 3 units^2 OR Approx 51.64 units^2
 
A(x) = (2x)(20-4x^2),see graph of A(x).

[attachment=0:9q11yoqs]mno.jpg[/attachment:9q11yoqs]
 

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Half of the area of the rectangle is (x)[f(x)].

The parabola crosses the x axis at -sqrt(5) and sqrt(5).
The vertical centreline of the rectangle is the y axis.

The "2x" that BigGlenn is referring to is twice the value of x between 0 and sqrt(5),
since the rectangle is twice the area of the part to the right of the y axis.

You could set length = L = 2x, for x ranging from 0 to sqrt(5) and H = f(x)
 
Ah. So i can't have my "length" cross the y-axis into the -X range of values. What i calculated then was from X=0 to X=sqrt(5/3) as my length and height up to the parabola. In essence i just got the area of the right half of the rectangle. Multiplying by 2 will get my total area of approx 103.28 units^2. (i'll get an exact answer when i get my calculator)
 
Hi Seekerofdragons,

When calculating area, you in general need to watch out for "x" being negative (or f(x) being so).
(When using integration, you must know the graph to avoid cancellation,
since integration does not calculate area! it multiplies f(x) by an interval,
f(x) is positive or negative.
Of course if f(x) is positive, no problem arises, you get area.)

You see, in calculating maximum area in this case, you correctly differentiated the Area function,
to locate the maximum point on the graph.
However, as the origin (0,0) is the midpoint of the rectangle base, if you use "x" for the base,
you are calculating the area of the right half of the rectangle, as you discovered.
You do not use the negative x, unless you make allowances for it being negative.
You are using x as the distance from the origin.

The reason this is happening is because the parabola is symmetric about the y-axis,
crossing the x axis at -sqrt(5) and sqrt(5).

I'm sorry I haven't got a way to upload graphs yet,
but if you draw the parabola (the area curve is a different parabola!),
you see the scenario of the rectangle being sliced in two by the y axis,
but you seem to see that quite easily.

Cheers.
 
some typos corrected, parabola crosses x axis at -sqrt(5) and sqrt(5).
 
Ok, on redoing my work with the corrections suggested earlier, I also found another error with my multiplication and found I was way, way off. apparently Sqrt(5/3)(40/3) != 40 Sqrt(15) / 3 should have been 40 Sqrt(15) / 9... go figure.

anyway, to double check my new answer and work regarding the area of a rectangle under a parabola:

Area = 2X(20-4X^2)
A(X) = 40X - 8X^3
A'(X) = 40 - 24X^2

X = Sqrt(15)/3

Area = 2(sqrt(15)/3) (20 - 4(sqrt(15)/3)^2)
Area = 2Sqrt(15)/3 (20 - 4(15/9))
Area = 2Sqrt(15)/3 (20 - 20/3)
Area = 2Sqrt(15)/3 (40/3)

Area = 80 sqrt(15)/9 ~ 34.4265 units^2
 
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