b The figure shows the cross section of a channel carrying water. Determine h, b and that minimize the length of the wetted perimeter while maintaining a cross- sectional area of 8 m². (Minimizing the wetted perimeter results in least resistance to the flow.) Ans: 14- Given. A = 8m² To Find: ktb h A Determine, h, b, 0 of wetted perimeter. h tano = m that minimize the length

b The figure shows the cross section of a channel carrying water. Determine h, b and that minimize the length of the wetted perimeter while maintaining a cross- sectional area of 8 m². (Minimizing the wetted perimeter results in least resistance to the flow.) Ans: 14- Given. A = 8m² To Find: ktb h A Determine, h, b, 0 of wetted perimeter. h tano = m that minimize the length

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Concept explainers

Rod Cutting

A naïve solution for the above-stated problem is to evaluate all of the possible combinations of different pieces of varying size to find the maximum revenue-generating combination. Although this method seems the easiest and quick to practice, it has exponential time complexity. So there is a need to figure out a better method than this.

Question

I've provided the written solution for the question. Could you please use the written solution to code the solution in python.

b
The figure shows the cross section of a channel carrying water. Determine h, b
and that minimize the length of the wetted perimeter while maintaining a cross-
sectional area of 8 m². (Minimizing the wetted perimeter results in least resistance to
the flow.)
Ans:
14-
KH
h
b
tant = m
Given.
A = 8m²
To Find:
Determine, h, b, 0 that minimize the length
of wetted perimeter.

As we
(1)
(ül)
A = √3
know, Face minimum perimeter (P)
2
= √3h²
(li) b = 2h
√3
y²
2
8 = √3 6²² + h = 2.15m.
And
b
tano =
T = 2 b
2=
tan o =
tano
2.48m.
=
2x2.15
53
(T-b)
h
(2b-b)
h
b
2h
0 = 29.97°
B
tan (25) =
tan
(b/₂).
h
2.48
2 X2.15

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