Optimization problems

  • Thread starter synergix
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  • #1
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Homework Statement


a) Show that of all the rectangles with a given area, the one with the smallest perimeter is a square.
b) Show that of all the rectangles with a given perimeter, the one with the greatest are is a square.


Homework Equations


As=x2
AR=xy
Ps = 4x
PR= 2x+2y

The Attempt at a Solution



I don't know how to do this any help?
 

Answers and Replies

  • #2
CompuChip
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Suppose that you have a rectangle with sides x and y.
So for a), the area A is fixed, which allows you to express y in terms of A and x.
Now the equation for the perimeter can be written in terms of x only, this will give you an equation which you can optimise for x.

The same is true for b, if you assume a fixed perimeter p. Perhaps b is even easier to start with, because you get the quadratic equation immediately.

Hint: what does the graph of f(x) = ax2 + bx + c look like and what is the x value of the maximum or minimum?

(That you will get the smallest perimeter in a but the largest area in b can also be seen from the formulas, by the way: when is the function value at the x that I asked you about in the hint a maximum? And when is it a minimum?).
 
  • #3
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for the rectangle

P=2x+2y
A=xy

y=A/x

p=2x+2A/x

take the derivative

p'=2+ (A'-xA)/x^2

solve for zero

2x^2 - Ax + A' = 0

x = [A +- sqrt(A^2 -8A')/4]

y= A / [A +- sqrt(A^2 -8A')/4]

P = 2[A +- sqrt(A^2 -8A')/4]+ A/[A +- sqrt(A^2 -8A')/4]


for square

P=4x
A=x^2

x= sqrt(A)

P=4sqrt(A)
 
  • #4
CompuChip
Science Advisor
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Note that the area of the rectangle is fixed as it is given. So A' = ... ?
This will simplify your equation greatly.

By the way, note that for a function
f(x) = a x2 + b x + c
the extremum (minimum or maximum, depending on the sign of a) is always at
x = - b / 2a
(following from f'(x) = 2ax + b = 0, or the fact that the extremum lies exactly between the two roots x = - b / 2a + sqrt(D) / 2a and x = - b / 2a - sqrt(D) / 2a of f(x)).

For the second one, note that the rectangular perimeter 2x + 2y is fixed to the value p. You don't know a priori that x = y, but what can you say about the area A = xy ?
 

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