Question

the process of frying food changes its quality, texture, and color. suppose the total change in color e (which is measured in the form of energy as kj/mol) of a cooked potato strip can be modeled by the function below, where c is the temperature (in °c) and t is the frying time (in min). complete parts a through c.
e(t,c)=433.56 - 10.57t - 5.44c - 0.02t² + 0.02c² + 0.08ct
a. what is the value of e prior to cooking? (assume that c = 0.)
e = 433.56 kj/mol
b. use this function to estimate the total change in color of a potato strip that has been cooked for 8 minutes at 150°c.
the total change in color is 77.72 kj/mol.
(type an integer or a decimal.)
c. determine the critical point of this function and determine if a maximum, minimum, or saddle point occurs at that point.
the critical point is at (t,c)=
(type an ordered pair, using integers or decimals. round to two decimal places as needed.)

Explanation:

Step1: Find the value of \(E\) prior to cooking

Given \(E(t,C)=433.56 - 10.57t-5.44C - 0.02t^{2}+0.02C^{2}+0.08Ct\) and \(C = 0\).
Substitute \(C = 0\) into the function:
\(E(t,0)=433.56-10.57t - 0.02t^{2}\)
When \(t = 0\) (prior to cooking), \(E(0,0)=433.56\) \(kJ/mol\)

Step2: Estimate the total change in color

We use the formula for the change of a function \(\Delta E\approx E_{t}(t,C)\Delta t+E_{C}(t,C)\Delta C\). Here \(t = 8\), \(C = 150\), \(\Delta t=0\) (since we are using the function at a single - point for estimation based on the given cooking time and temperature, and we assume small - step approximation conceptually, but if we consider the total change from \(t = 0,C = 0\) to \(t = 8,C = 150\) using the function values:
First, find \(E(8,150)\):

$$
LATEXBLOCK0
$$

Step3: Find the critical point

We find the first - order partial derivatives.
\(E_{t}(t,C)=-10.57-0.04t + 0.08C\)
\(E_{C}(t,C)=-5.44 + 0.04C+0.08t\)
Set \(E_{t}(t,C)=0\) and \(E_{C}(t,C)=0\)
From \(E_{t}(t,C)=0\), we have \(-10.57-0.04t + 0.08C=0\), which can be rewritten as \(0.08C-0.04t=10.57\) or \(2C - t=\frac{10.57}{0.02}=528.5\) (Equation 1)
From \(E_{C}(t,C)=0\), we have \(-5.44 + 0.04C+0.08t=0\), which can be rewritten as \(0.04C+0.08t=5.44\) or \(C + 2t=\frac{5.44}{0.04}=136\) (Equation 2)
Multiply Equation 2 by 2: \(2C + 4t=272\) (Equation 3)
Subtract Equation 1 from Equation 3:
\((2C + 4t)-(2C - t)=272 - 528.5\)
\(5t=-256.5\)
\(t=- 51.3\)
Substitute \(t=-51.3\) into Equation 2: \(C+2\times(-51.3)=136\)
\(C=136 + 102.6=238.6\)

Answer:

a. \(433.56\) \(kJ/mol\)
b. \(77.72\) \(kJ/mol\)
c. \((t,C)=(-51.30,238.60)\)