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 blanched potato strips can be estimated 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
(type an integer or a decimal.)
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) = (1.55,132.90).
(type an ordered pair, using integers or decimals. round to two decimal places as needed.)
what kind of point is the critical point?
a. a relative maximum
b. a relative minimum
c. a saddle point
d. the kind of point cannot be determined.

Explanation:

Step1: Find the value of E when \(C = 0\)

Substitute \(C = 0\) into the function \(E(t,C)=433.56 - 10.57t-5.44C - 0.02t^{2}+0.02C^{2}+0.08Ct\).
When \(C = 0\), the function becomes \(E(t,0)=433.56-10.57t - 0.02t^{2}\).
Since we are not given a value of \(t\) for part (a), and if we assume \(t = 0\) (prior to cooking, \(t = 0\) minutes).
Substitute \(t = 0\) into \(E(t,0)\):
\(E(0,0)=433.56-10.57\times0 - 0.02\times0^{2}=433.56\)

Step2: Second - derivative test for critical points

For a function \(z = f(x,y)\) (here \(x=t\), \(y = C\)), the second - derivative test uses \(D=f_{tt}f_{CC}-(f_{tC})^{2}\)
First, find the first - order partial derivatives:
\(E_{t}=\frac{\partial E}{\partial t}=-10.57 - 0.04t+0.08C\)
\(E_{C}=\frac{\partial E}{\partial C}=-5.44 + 0.04C+0.08t\)
Then find the second - order partial derivatives:
\(E_{tt}=\frac{\partial^{2}E}{\partial t^{2}}=- 0.04\)
\(E_{CC}=\frac{\partial^{2}E}{\partial C^{2}}=0.04\)
\(E_{tC}=\frac{\partial^{2}E}{\partial t\partial C}=0.08\)

Calculate \(D = E_{tt}E_{CC}-(E_{tC})^{2}\)
\(D=(-0.04)\times(0.04)-(0.08)^{2}\)
\(D=-0.0016 - 0.0064=-0.008<0\)

Answer:

a. \(E = 433.56\) kJ/mol
b. The total change in color is \(77.72\) kJ/mol
c. Since \(D=-0.008<0\), the critical point \((1.55,132.90)\) is a saddle point. So the answer is B. A saddle point.