the process of frying food changes its quality, texture, and color. suppose the total change in color e…

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.\ne(t,c) = 433.56 - 10.57t - 5.44c - 0.02t² + 0.02c² + 0.08ct\na. what is the value of e prior to cooking? (assume that c = 0.)\ne = 433.56 kj/mol\n(type an integer or a decimal.)\nb. use this function to estimate the total change in color of a potato strip that has been cooked for 8 minutes at 150°c.\nthe total change in color is 77.72 kj/mol.\n(type an integer or a decimal.)\nc. determine the critical point of this function and determine if a maximum, minimum, or saddle point occurs at that point.\nthe critical point is at (t,c) = (1.55,132.90).\n(type an ordered pair, using integers or decimals. round to two decimal places as needed.)\nwhat kind of point is the critical point?\na. a relative maximum\nb. a relative minimum\nc. a saddle point\nd. the kind of point cannot be determined.
Answer
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.