Classic optimization: wire, window, fence practice problems
The classic optimization problems: an open box from a sheet, a Norman window, a can, a cut wire. 25 problems with worked solutions; in 25 of them every
equation is proved by a computer algebra system.
4 worked problems
Try each one before opening the solution. Every line is labeled with what checked it. How we verify
A Norman window is a rectangle topped by a semicircle. If its perimeter is \( \displaystyle 8 \) ft, what radius \( \displaystyle x \) of the semicircle lets in the most light (the largest area), and what is that area?
Width 2x, height h = (8 − 2x − πx)/2 from the perimeter; A(x) = 2x·h + πx²/2.
Reviewed
\[ \frac{d}{d x} \left(\frac{\pi x^{2}}{2} + 2 x \left(- \frac{\pi x}{2} - x + 4\right)\right) = - 4 x - \pi x + 8 \]
Differentiate.✓ Proved
\[ \left. - 4 x - \pi x + 8 \right|_{\substack{ x=\frac{8}{\pi + 4} }} = 0 \]
The only critical point in the domain is x = 8/(pi + 4).✓ Proved
✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
A wire \( \displaystyle 10 \) m long is cut in two. One piece (length \( \displaystyle x \)) is bent into a square, the other into a circle. What \( \displaystyle x \) makes the total area as small as possible, and what is that area?
Square side x/4; circle circumference 10 − x, radius (10 − x)/(2π). A(x) = (x/4)² + (10 − x)²/(4π).
Reviewed
\[ \frac{d}{d x} \left(\frac{x^{2}}{16} + \frac{\left(10 - x\right)^{2}}{4 \pi}\right) = \frac{\pi x + 4 x - 40}{8 \pi} \]
Differentiate.✓ Proved
\[ \left. \frac{\pi x + 4 x - 40}{8 \pi} \right|_{\substack{ x=\frac{40}{\pi + 4} }} = 0 \]
The only critical point in the domain is x = 40/(pi + 4).✓ Proved
✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
A closed cylindrical can must hold \( \displaystyle 16 \pi \) cubic inches. What radius \( \displaystyle x \) uses the least material, and what is that surface area?
h = 16*pi/(πx²) from the volume, so S(x) = 2πx² + 2πx·h = 2πx² + 32π/x.
✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
An open box is made from a \( \displaystyle 36 \times 36 \) sheet by cutting equal squares of side \( \displaystyle x \) from the corners and folding up the sides. What \( \displaystyle x \) gives the largest volume, and what is it?
V(x) = x(36 − 2x)², for 0 < x < 18.
Reviewed
\[ \frac{d}{d x} x \left(36 - 2 x\right)^{2} = \left(x - 6\right) \left(12 x - 216\right) \]
✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
More classic optimization: wire, window, fence practice