Practice Dimensions of Shortest Diagonal

puremath

New member
Among all rectangles with a given perimeter P, find the dimensions of the one with the shortest diagonal.

Seeking steps or set up. Thanks.
 

MarkFL

La Villa Strangiato
Math Helper
Let's let the rectangle have width \(w\) and height \(h\). And so our objective function is:

[MATH]d=\sqrt{w^2+h^2}[/MATH]
We have the constraint:

[MATH]2(w+h)=P[/MATH]
Use the constraint to write the objective function in terms of one variable, which will give you a quadratic radicand.
 

puremath

New member
Let's let the rectangle have width \(w\) and height \(h\). And so our objective function is:

[MATH]d=\sqrt{w^2+h^2}[/MATH]
We have the constraint:

[MATH]2(w+h)=P[/MATH]
Use the constraint to write the objective function in terms of one variable, which will give you a quadratic radicand.

Are you saying to solve 2w + 2h = P for either h or w?
 

MarkFL

La Villa Strangiato
Math Helper
Yes.
 

puremath

New member

2w + 2h = P

2w = P - 2h

w = (P - 2h)/2

Do I now plug w into d = sqrt{w^2 + h^2}? If so, what is the reason for this substitution?
 

MarkFL

La Villa Strangiato
Math Helper
Yes. We want to have a function in one variable.
 

MarkFL

La Villa Strangiato
Math Helper
Here is a live graph to illustrate the problem:


Move the slider for \(w\) to change the width of the rectangle. It has a fixed perimeter of 1 unit. From this graph it is easy to see that this problem is mathematically equivalent to finding the point on the constraint:

[MATH]2x+2y=P[/MATH]
that is closest to the origin. We know this point must lie along the line:

[MATH]y=x[/MATH]
(How do we know this?)

And so this means the diagonal is minimized when \(w=h\), which means the rectangle is a square. Thus, from this and another recent problem, we know that given a rectangle with a given fixed perimeter, the area is maximized when it's a square, and the diagonal is simultaneously minimized.
 

puremath

New member
Here is a live graph to illustrate the problem:


Move the slider for \(w\) to change the width of the rectangle. It has a fixed perimeter of 1 unit. From this graph it is easy to see that this problem is mathematically equivalent to finding the point on the constraint:

[MATH]2x+2y=P[/MATH]
that is closest to the origin. We know this point must lie along the line:

[MATH]y=x[/MATH]
(How do we know this?)

The point must be on the line y = x because this line passes through the origin.

And so this means the diagonal is minimized when \(w=h\), which means the rectangle is a square. Thus, from this and another recent problem, we know that given a rectangle with a given fixed perimeter, the area is maximized when it's a square, and the diagonal is simultaneously minimized.
 

MarkFL

La Villa Strangiato
Math Helper
To follow up, we may write the square of the diagonal as:

[MATH]d^2=w^2+\left(\frac{P}{2}-w\right)^2=2w^2-Pw+\frac{P^2}{4}[/MATH]
The axis of symmetry is:

[MATH]w=\frac{P}{4}\implies h=w=\frac{P}{4}[/MATH]
 

puremath

New member
To follow up, we may write the square of the diagonal as:

[MATH]d^2=w^2+\left(\frac{P}{2}-w\right)^2=2w^2-Pw+\frac{P^2}{4}[/MATH]
The axis of symmetry is:

[MATH]w=\frac{P}{4}\implies h=w=\frac{P}{4}[/MATH]

Very cool.
 
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