What Dimensions Maximize the Area of a Rectangle Under y = 8 - x^3?

In summary: But in summary, the largest rectangle that can be drawn in the closed region bounded by the x- and y-axes and the graph of the function y = 8 - x3 has dimensions of 15. However, there is no further explanation or clarification provided in the conversation or in the answer.
  • #1
Procrastinate
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What are the dimensions of the largest rectangle that can be drawn in the closed region bounded by the x- and y-axes and the graph of the function y = 8 - x3
A=lw=16
D= 2l + 2w = 2l + 16/l

I started off by graphing the equation. The y-int was 8 and the x-int was 2.
Thus, I established that the total area bound by the axes was 16.
Afterwards, I found the stationary points of D:

D'=2-32/l2
D''=-64/l2
The stationary point was four and thus as D'' was a negative, D is a maximum. L=4 and thus w=4 and the total dimensions were 16.


However, this was wrong, the answer was 15. I think I may have interpreted the question incorrectly but could someone please point out my error? Thank you.
 
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  • #2
I think I may have interpreted the question incorrectly also... I first assumed "the largest rectangle" to be the one with most area. From what you've done I think you're assuming it's the largest perimeter.

I started off by graphing the equation. The y-int was 8 and the x-int was 2.
Thus, I established that the total area bound by the axes was 16.
The total area bounded by the axes and the function is not 2x8=16, it's [tex]\int_0^2ydx[/tex] where [itex]y=8-x^3[/itex]

For the largest area, I found an answer of 6.21/3 which have dimensions x=21/3 and y=6. For the largest perimeter, I have x=1/31/2.

So I'm at a complete loss of how the book gets the answer of 15...
 
  • #3
P = 2x + 2y
A = xy
Maximize P.


I think you can only get 15 as the answer if you find maximum perimeter here. Something like x~1 and y~6.5., P~2+13=15

Although x=1 and y=7 give you greater perimeter? I don't get how the book gets 15! Maybe he rotated the rectangle like 45 degrees.. :confused:
 
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  • #4
shouldnt you maximize area?
 
  • #5
Procrastinate, can you please be more clear on what the answer from the book is. "15" on its own doesn't tell us much, and makes things especially confusing when the question itself can be interpreted in a thousand different ways.
 
  • #6
Mentallic said:
Procrastinate, can you please be more clear on what the answer from the book is. "15" on its own doesn't tell us much, and makes things especially confusing when the question itself can be interpreted in a thousand different ways.

The answer just says 15 and doesn't provide any further explanation, unfortunately.
 
  • #7
Ok well can you at least tell us what your mathematical background is? If the answer is looking for a rotated rectangle as cronxeh suggested, then this is obviously going to be a lot more complicated than the average calculus 1 student can do.
 
  • #8
And can I just point out something,

Procrastinate said:
A=lw=16
D= 2l + 2w = 2l + 16/l

w=16/l, therefore D=2l+2(16/l)=2l+32/l

not what you have.
 
  • #9
High school maths student.
 
  • #10
Then it's safe to say that we are restrained to the rectangle being "parallel" to the axes. i.e. not turned at some angle.

Well, this question still bothers me...

book said:
Question:What are the dimensions of the largest rectangle that can be drawn in the closed region bounded by the x- and y-axes and the graph of the function y = 8 - x3

Answer: 15

wow...
 

FAQ: What Dimensions Maximize the Area of a Rectangle Under y = 8 - x^3?

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