# PracticeDimensions 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

#### 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?

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.

• puremath

#### 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

#### 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

#### 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.