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- Thread starter puremath
- Start date

[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.

[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?

Yes.

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?

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.

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.

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.