Explain why this is correct (Optimization Problem)

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SUMMARY

The optimization problem involves determining the dimensions of a rectangle with a maximum area using a 100 cm wire. The perimeter equation is defined as P = 2(l + w), leading to the relationship w = (100 - 2l)/2. The area A is expressed as A = lw, which simplifies to A = 50l - l² when substituting for w. The critical point is found by setting the derivative A' = 50 - l = 0, yielding l = 25 cm and w = 25 cm, confirming that the rectangle is a square for maximum area.

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  • Understanding of basic algebraic manipulation
  • Familiarity with optimization concepts in calculus
  • Knowledge of perimeter and area formulas for rectangles
  • Ability to differentiate functions
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  • Explore the relationship between perimeter and area in geometric shapes
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IbrahimA
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Homework Statement


A piece of wire, 100 cm long, needs to be bent to form a rectangle. Determine the dimensions of a rectangle with the maximum area.

Homework Equations


P = 2(l+w)
A = lw

The Attempt at a Solution


This is what I don't understand, the solutions that I saw from looking around is:

l = x
w = (100 -2x) / 2

I don't understand why is width portrayed as shown above, and why the length is also potrayed as above, the solution goes onto:

A = (x)(100 - 2x / 2)
A = (x)(50 = x)
A = 50x - x^2
A prime = 50 - x^2
Insert 0 for A prime
0 = 50 - x^2
x=25
With therefore means the length and width are 25cm.

I understand the algebra, I do not understand how to get the length and width equation, wondering if someone could explain it to me.
 
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IbrahimA said:

Homework Statement


A piece of wire, 100 cm long, needs to be bent to form a rectangle. Determine the dimensions of a rectangle with the maximum area.

Homework Equations


P = 2(l+w)
A = lw

The Attempt at a Solution


This is what I don't understand, the solutions that I saw from looking around is:

l = x
w = (100 -2x) / 2

I don't understand why is width portrayed as shown above, and why the length is also potrayed as above,
The 100 cm of wire will be the perimeter of the rectangle. From your formula for P, solve for the width w in terms of l. Also, introducing x as a variable is more complicating than just using w and l.
[IbrahimA said:
the solution goes onto:

A = (x)(100 - 2x / 2)
A = (x)(50 = x)
The first equation above is not written correctly. The second group in parentheses is 100 - 2x/2, which is properly interpreted as ##100 - \frac {2x} 2= 100 - x##. Obviously that's not what you meant, so the equation should have been written as A = x(100 - 2x)/2.

The second equation has a typo -- you wrote = instead of -.
[IbrahimA said:
A = 50x - x^2
A prime = 50 - x^2
Insert 0 for A prime
0 = 50 - x^2
x=25
With therefore means the length and width are 25cm.

I understand the algebra, I do not understand how to get the length and width equation, wondering if someone could explain it to me.
 
Forget about the characteristic of length and width... a rectangle has two pairs of parallel sides... one pair has length x and the other has length y. Note the special case that x=y is allowed even though we said "rectangle".
If y>x then y is the length. If x>y then x is the length... but it does not matter to the maths, we can focus on either.

At this stage the x and y aee just labels... the next step uses maths to describe how these are related to the area and the perimeter. If the area is A and the perimeter is p, write down the equations for these in terms of x and y.

You need an ewuation for area in terms of only one other variable... so pick x or y, doesn't matter which, and make A depend only on that.

Note. Your final working contains two errors which cancel each other out... check the derivative.
 
Last edited:

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