/*! 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 19 Cooking Loss Milk proteins are s... [FREE SOLUTION] | 91Ó°ÊÓ

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Cooking Loss Milk proteins are sometimes added to sausage to reduce shrinkage due to cooking loss and to improve the texture of the sausage. Research has shown that when sausage is prepared with three milk proteins-sodium caseinate, whey protein, and skim milk powder-the cooking loss (expressed as a percentage of initial weight) can be modeled as \(L(w, s)=10.65+1.13 w+1.04 s-5.83 w s\) percent where \(w\) is the proportion of whey protein, \(s\) is the proportion of skim milk powder. a. Locate the point where the partial derivatives of \(L\) are both equal to zero. b. What type of critical point is the point in part \(a\) ?

Short Answer

Expert verified
(w, s) = (0.1784, 0.1938); it's a saddle point.

Step by step solution

01

Find Partial Derivatives

First, we need to calculate the partial derivatives of the function \(L(w, s) = 10.65 + 1.13w + 1.04s - 5.83ws\). The partial derivative with respect to \(w\) is: \(\frac{\partial L}{\partial w} = 1.13 - 5.83s\).The partial derivative with respect to \(s\) is: \(\frac{\partial L}{\partial s} = 1.04 - 5.83w\).
02

Set Partial Derivatives to Zero

Set each of these partial derivatives equal to zero to find the critical points:1. \(\frac{\partial L}{\partial w} = 0\): \[1.13 - 5.83s = 0\] \[s = \frac{1.13}{5.83}\].2. \(\frac{\partial L}{\partial s} = 0\): \[1.04 - 5.83w = 0\] \[w = \frac{1.04}{5.83}\].
03

Solve for w and s

Substitute the values found in Step 2:- \(s = \frac{1.13}{5.83} \approx 0.1938\)- \(w = \frac{1.04}{5.83} \approx 0.1784\)The critical point where both partial derivatives are zero is approximately \((w, s) = (0.1784, 0.1938)\).
04

Determine the Type of Critical Point

To determine the type of critical point, calculate the second partial derivatives:- \(\frac{\partial^2 L}{\partial w^2} = 0\)- \(\frac{\partial^2 L}{\partial s^2} = 0\)- \(\frac{\partial^2 L}{\partial w \partial s} = -5.83\).The Hessian determinant is \[ D = \frac{\partial^2 L}{\partial w^2} \cdot \frac{\partial^2 L}{\partial s^2} - \left(\frac{\partial^2 L}{\partial w\partial s}\right)^2 = 0 \cdot 0 - (-5.83)^2 < 0\].Since \(D < 0\), the critical point is a saddle point.

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Key Concepts

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

Critical Points
In multivariable calculus, finding critical points is an essential step in solving optimization problems. Critical points occur where the partial derivatives of a function are zero. These points can give us information about the function's behavior, such as finding local maxima, minima, or saddle points.

For a given function, in our scenario \( L(w, s) = 10.65 + 1.13w + 1.04s - 5.83ws \), we compute the partial derivatives with respect to both variables. These derivatives help us understand how the function \( L \) changes with small variations in \( w \) and \( s \).
  • The partial derivative with respect to \( w \) is \( \frac{\partial L}{\partial w} = 1.13 - 5.83s \).
  • The partial derivative with respect to \( s \) is \( \frac{\partial L}{\partial s} = 1.04 - 5.83w \).
By setting each of these derivatives to zero, we find the point \( (w, s) \) where both \( \frac{\partial L}{\partial w} \) and \( \frac{\partial L}{\partial s} \) are zero, resulting in a critical point. Solving these equations gives us the approximate values \( w = 0.1784 \) and \( s = 0.1938 \). This critical point is crucial for further analysis to determine the nature of this point.
Saddle Point
A saddle point is a type of critical point in a function of two variables where the function doesn't have a local maximum or minimum. Instead, it sits on a surface that curves upwards in one direction and downwards in another.

In our function \( L(w, s) = 10.65 + 1.13w + 1.04s - 5.83ws \), once we've determined the critical point at \( (w, s) = (0.1784, 0.1938) \), assessing whether this point is a saddle point requires additional information beyond just finding the partial derivatives. The concept of a saddle point means that around this critical point, the surface of \( L \) appears "saddle-like."
  • In one direction, if you move, you see an increase in the function's value, suggesting a peak.
  • In another direction, the function's value decreases, suggesting a dip.
This behavior indicates that the critical point isn't a local extremum, but a saddle point. The determination whether or not a critical point is a saddle uses further calculations with the Hessian determinant, as we'll see in the following section.
Hessian Determinant
The Hessian determinant is a key determinant in multivariable calculus that helps establish the nature of a critical point detected by setting partial derivatives to zero. It is derived from the second partial derivatives of the function.

For the function \( L(w, s) \), second derivatives we calculate included:
  • \( \frac{\partial^2 L}{\partial w^2} = 0 \)
  • \( \frac{\partial^2 L}{\partial s^2} = 0 \)
  • \( \frac{\partial^2 L}{\partial w \partial s} = -5.83 \)
The Hessian determinant \( D \) is computed as:\[D = \frac{\partial^2 L}{\partial w^2} \cdot \frac{\partial^2 L}{\partial s^2} - \left( \frac{\partial^2 L}{\partial w \partial s} \right)^2\]Calculating this for our function:\[D = 0 \cdot 0 - (-5.83)^2 = -34.0129\]Since \( D < 0 \), this indicates the critical point \( (0.1784, 0.1938) \) is indeed a saddle point. When the determinant is negative, it implies that the critical point is not a maximum or minimum but rather a point of inflection-like behavior, characteristic of saddle points.

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Most popular questions from this chapter

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