/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 Consider the first-order model$$... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the first-order model$$y=12+1.2 x_{1}-2.1 x_{2}+1.6 x_{3}-0.6 x_{4}$$ where \(-1 \leq x_{i} \leq 1\) a. Find the direction of steepest ascent. b. Assume that the current design is centered at the point \((0,0,0,0) .\) Determine the point that is three units from the current center point in the direction of steepest ascent.

Short Answer

Expert verified
Direction: (1.2, -2.1, 1.6, -0.6); Point: (1.216, -2.128, 1.622, -0.608).

Step by step solution

01

Identify Partial Derivatives

The direction of steepest ascent is determined by the vector of partial derivatives (the gradient vector) of the function with respect to each variable \(x_i\). For our function \(y = 12 + 1.2x_1 - 2.1x_2 + 1.6x_3 - 0.6x_4\), calculate the partial derivatives: \[ \frac{\partial y}{\partial x_1} = 1.2, \quad \frac{\partial y}{\partial x_2} = -2.1, \quad \frac{\partial y}{\partial x_3} = 1.6, \quad \frac{\partial y}{\partial x_4} = -0.6 \]
02

Formulate Gradient Vector

The gradient vector, which gives the direction of steepest ascent, is formed from these partial derivatives: \( abla y = (1.2, -2.1, 1.6, -0.6) \).
03

Normalize the Gradient Vector

To determine the unit direction vector, normalize the gradient vector. Calculate the magnitude of the gradient vector: \[ \| abla y \| = \sqrt{1.2^2 + (-2.1)^2 + 1.6^2 + (-0.6)^2} = \sqrt{1.44 + 4.41 + 2.56 + 0.36} = \sqrt{8.77} \approx 2.96 \] The unit vector in the direction of steepest ascent is \( \left( \frac{1.2}{2.96}, \frac{-2.1}{2.96}, \frac{1.6}{2.96}, \frac{-0.6}{2.96} \right) \).
04

Calculate Point in Direction of Steepest Ascent

Starting from the point \((0,0,0,0)\), to move three units in the direction of the unit gradient vector, multiply each component of the unit vector by 3: \[ 3 \left( \frac{1.2}{2.96}, \frac{-2.1}{2.96}, \frac{1.6}{2.96}, \frac{-0.6}{2.96} \right) = \left( 3 \times \frac{1.2}{2.96}, 3 \times \frac{-2.1}{2.96}, 3 \times \frac{1.6}{2.96}, 3 \times \frac{-0.6}{2.96} \right) \] Calculating gives: \( (1.216, -2.128, 1.622, -0.608) \).
05

Conclude with Final Direction and Point

The direction of steepest ascent is given by the gradient vector \((1.2, -2.1, 1.6, -0.6)\). The point three units in this direction from \((0,0,0,0)\) is \((1.216, -2.128, 1.622, -0.608)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gradient Vector
The gradient vector is a mathematical tool used to find the direction of steepest ascent in multi-variable calculus. In simpler terms, it helps in determining the direction in which a function increases most rapidly. To calculate the gradient vector, we first need to find the partial derivatives of the function with respect to each variable. Each of these derivatives represents a component of the gradient vector.
For instance, in the given model, the partial derivatives are used to form the gradient vector, denoted as \( abla y = (1.2, -2.1, 1.6, -0.6) \). Each element in this vector represents the rate of change of the function along one of the variables \(x_1, x_2, x_3,\) and \(x_4\). The sign and magnitude of these components can give us insights into how the function behaves in different directions.
Partial Derivatives
Partial derivatives are fundamental in understanding how a function changes as each input variable is varied while keeping others constant. They are the building blocks of the gradient vector and play a crucial role in multivariable calculus.
In our exercise, the partial derivatives were calculated as follows:
  • \( \frac{\partial y}{\partial x_1} = 1.2 \)
  • \( \frac{\partial y}{\partial x_2} = -2.1 \)
  • \( \frac{\partial y}{\partial x_3} = 1.6 \)
  • \( \frac{\partial y}{\partial x_4} = -0.6 \)
These values, derived from the coefficients of each variable in the function, indicate how much the function \(y\) will change if we slightly increase or decrease the corresponding \(x_i\) while keeping all other variables unchanged.
Unit Direction Vector
A unit direction vector is essentially the gradient vector adjusted to have a magnitude (or length) of one. This vector indicates the direction of steepest ascent without concern for scale, making it easier to work with in applications where direction rather than magnitude is important.
To obtain this, we take the gradient vector and divide each of its components by its magnitude. In our exercise, the magnitude of the gradient vector \( abla y = (1.2, -2.1, 1.6, -0.6) \) was calculated to be approximately 2.96. So, the unit direction vector becomes \( \left( \frac{1.2}{2.96}, \frac{-2.1}{2.96}, \frac{1.6}{2.96}, \frac{-0.6}{2.96} \right) \).
This adjustment ensures we direct our movement proportionally in the steepest ascent direction without altering the overall scale.
Normalization of Gradient
Normalization is the process used to convert any vector into a unit vector. A unit vector preserves direction but ensures the length equals one. This process simplifies calculus and optimization tasks like finding a specific point from a given start point.
To normalize the gradient vector, compute its magnitude: \[ \| abla y \| = \sqrt{1.2^2 + (-2.1)^2 + 1.6^2 + (-0.6)^2} = \sqrt{8.77} \approx 2.96 \]Next, divide each element of the gradient vector by this magnitude:\[ \left( \frac{1.2}{2.96}, \frac{-2.1}{2.96}, \frac{1.6}{2.96}, \frac{-0.6}{2.96} \right) \]This forms a unit vector, maintaining direction but adjusting magnitude. We use the unit vector to navigate precise steps in the problem-solving process, like moving a set distance from a center point.

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