Optimizing Rectangle Dimensions for Area 1000m^2: A Simple Solution

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To find the dimensions of a rectangle with an area of 1000 m² and the smallest perimeter, the perimeter can be expressed as 2x + 2000/x. Differentiating this equation helps identify the minimum perimeter. The optimal dimensions are derived by setting the derivative equal to zero, leading to x being the square root of 1000, or approximately 31.62 m. Consequently, the corresponding y dimension can be calculated as 1000/x, resulting in a rectangle with dimensions that minimize the perimeter effectively.
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[SOLVED] simple optimization problem

Homework Statement



find the dimensions of a rectangle with area 1000 m^2 whose perimeter is as small as possible

Homework Equations



perimeter = 2x + 2y

area = xy

1000 = xy

y= 1000/x

perimeter = 2x + 2(1000/x)



The Attempt at a Solution



am stuck

any help from anybody?
 
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perimeter = 2x + 2(1000/x)
now differentiate!

should learn concepts from the book (if you don't know why you differentiate)

Simple approach:

plot your perimeter and pick and minimum value
this is realistic problem, so going for really large x and -ve x would be nonsense
 
2-(2000/x^2)=0

i think
 
physicsed said:
2-(2000/x^2)=0

i think
You think? ... have confidence!
 
2/1000=x^-2
 
no, try 2=2000/x^2
 
x=square root of .001?
 
x=\sqrt{1000}=\sqrt{10\cdot10^2}=10\sqrt{10}
 
x=\sqrt{1000}=\sqrt{10\cdot10^2}=10\sqrt{10}

how did u get that?
 
  • #10
2=\frac{2000}{x^2}\rightarrow x^2=\frac{2000}{2}\rightarrow x^2=1000
 
  • #11
thanks alot. am messed up today!
 

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