min area of a rectangle with a given function

[lisamaee]

I've been given the following problem:

Consider rectangles with an edge on the x-axis and two vertices (=corners) above the x-axis on the curve y=4-x^2. Among such rectangles, is there one with a maximum area? Is there one with a minimum area? What are the dimensions?

I have already solved for the max by doing the following:

A=xy
and because y=4-x^2
A=x(4-x^2) =4x-x^3
dA/dx=4-3x^2 ...... x=2/sqrt(3)
y=4-x^2 =4-(4/3) =8/3
since A=xy ... A=16/(3sqrt(3))

I don't understand how the steps to finding the min area differ. Help please, my final is tomorrow morning! :/

 

[DrPhil]

I've been given the following problem:

Consider rectangles with an edge on the x-axis and two vertices (=corners) above the x-axis on the curve y=4-x^2. Among such rectangles, is there one with a maximum area? Is there one with a minimum area? What are the dimensions?

I have already solved for the max by doing the following:

A=xy
and because y=4-x^2
A=x(4-x^2) =4x-x^3
dA/dx=4-3x^2 ...... x=2/sqrt(3)
y=4-x^2 =4-(4/3) =8/3
since A=xy ... A=16/(3sqrt(3))

I don't understand how the steps to finding the min area differ. Help please, my final is tomorrow morning! :/

You should show the crucial step pf setting dA/dx = 0, explicitly. Remember that tells you the relative min/max within the domain. To find absolute min/max, you must also ask what are the values at the limits of the domain. What happens if x=0, or if x=+/-2 ?

 

[lisamaee]

You should show the crucial step pf setting dA/dx = 0, explicitly. Remember that tells you the relative min/max within the domain. To find absolute min/max, you must also ask what are the values at the limits of the domain. What happens if x=0, or if x=+/-2 ?

dA/dx=0 is in my original work, I just didn't type it out (hence the .....)

But I'm still confused, is the minimum area 0 for when x=+and-2? Can a line count as a rectangle?

 

[DrPhil]

dA/dx=0 is in my original work, I just didn't type it out (hence the .....)

But I'm still confused, is the minimum area 0 for when x=+and-2? Can a line count as a rectangle?

Yes indeed, a rectangle of zero width when x=0, and a rectangle of zero height when x=+/-2 If either h or w is 0, the Area is 0.

 

[lookagain]

Yes indeed, a rectangle of zero width when x=0, and a rectangle of zero height when x=+/-2
If either h or w is 0, the Area is 0.

Then they cease to be rectangles, because rectangles necessarily have positive areas.**
Polygons have interiors. Line segments have no interiors.


Is there one (a rectangle with those conditions) with a minimum area? No.




** I don't mean positive area in the sense of the direction with which its
vertices are taken when figuring their areas.

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Edit:

Rectangles have four sides. They have opposite sides, interior angles, and exterior
angles. They have interiors and exteriors. The interiors do not have zero area.

Line segments have none of the above.

 

[DrPhil]

Rectangles have four sides. They have opposite sides, interior angles, and exterior
angles. They have interiors and exteriors. The interiors do not have zero area.

Line segments have none of the above.

Looking at the formula lisamaee derived for A(x),

A(x) = x(4 - x^2) = 4x - x^3 ,

the area is zero if x is -2, 0, or +2. The question is whether those points are within the domain of x, or is the domain limited to open intervals, (-2,0) and (0,+2).

As a practical user of calculus (not a professional mathematician), I am not bothered by allowing a "rectangle" to have zero area. With either definition, it is a rectangle as x approaches any of these three points. For any positive value of delta, no matter how small, we can find epsilon such that A(epsilon) < delta. The limit of A(x) = 0, so I would say the minimum value of A is 0.

 

[MarkFL]

I've been given the following problem:

Consider rectangles with an edge on the x-axis and two vertices (=corners) above the x-axis on the curve y=4-x^2. Among such rectangles, is there one with a maximum area? Is there one with a minimum area? What are the dimensions?

I have already solved for the max by doing the following:

A=xy
and because y=4-x^2
A=x(4-x^2) =4x-x^3
dA/dx=4-3x^2 ...... x=2/sqrt(3)
y=4-x^2 =4-(4/3) =8/3
since A=xy ... A=16/(3sqrt(3))

I don't understand how the steps to finding the min area differ. Help please, my final is tomorrow morning! :/

You want \(\displaystyle A=2xy\)... since the base of the rectangle is \(\displaystyle x-(-x)=2x\) and the height is \(\displaystyle y\).