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Problem 3 modelling

Food safety precautions state that, when a certain food is being reheated, it should either:

reach a temperature of above $75^{\circ} C$ for at least $30$ seconds
or, reach a maximum core temperature of at least $80^{\circ} C$

in order to make it safe to eat.

An industrial kitchen models a reheating process using the formula

T = 45 + 2 t - \frac{1}{32} t^{2}, 0 \leq t

where $T$ is the core temperature of the food product, and $t$ is time in seconds.

Determine whether this process should be considered safe, or whether the process needs to be altered, explaining your answer.
Let $δ$ equal the duration for which the temperature of the product is above $75^{\circ} C$ , and let the maximum temperature of the product be $θ .$

Give the value of $δ θ$ .

Hint

To see if the first criterion is satisfied, we are interested in solving the inequality

45 + 2 t - \frac{1}{32} t^{2} > 75

Hint

To see if the second criterion is satisfied, we need to find the maximum value of

T = 45 + 2 t - \frac{1}{32} t^{2}

We can do this by completing the square, or by considering the symmetry of the parabola.

Solution

To see if the first criterion is satisfied, we are interested in solving the inequality

45 + 2 t - \frac{1}{32} t^{2} > 75

We find the roots first

\begin{aligned} 45 + 2 t - \frac{1}{32} t^{2} & = 75 \\ - \frac{1}{32} t^{2} + 2 t - 30 & = 0 \\ t^{2} - 64 t + 960 & = 0 \\ (t - 24) (t - 40) & = 0 \\ t & \in {24, 40} \end{aligned}

Let's make a sketch:

rectangle

Looking at the sketch, we see that the temperature is above $75$ for $40 - 24 = 16$ seconds.

By symmetry, the maximum temperature would occur at the midpoint of $24$ and $40$ , which is $t = 32$ , giving

\begin{aligned} T & = 45 + 2 (32) - \frac{1}{32} (32)^{2} \\ = 77 \end{aligned}

Since the duration above $75$ is less than $30$ seconds, and the maximum temperature is less than $80$ , the process is not considered safe and needs to be altered.

Problem 3 modelling ​

Problem 3 modelling