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Quality Control of Cheese It is often difficult to
Chapter 3, Problem 57A(choose chapter or problem)
Quality Control of Cheese It is often difficult to evaluate the quality of products that undergo a ripening or maturation process. Researchers have successfully used ultrasonic velocity to determine the maturation time of Mahon cheese. The age can be determined by
\(M(v)=0.0312443 v^{2}-101.39 v+82,264, \quad v \geq 1620\),
where M(v) is the estimated age of the cheese (in days) for a velocity v (m per second). Source: Journal of Food Science.
a. If Mahon cheese ripens in 150 days, determine the velocity of the ultrasound that one would expect to measure. (Hint: Set M(v) = 150 and solve for v.)
b. Determine the derivative of this function when v = 1700 m per second and interpret.
Questions & Answers
QUESTION:
Quality Control of Cheese It is often difficult to evaluate the quality of products that undergo a ripening or maturation process. Researchers have successfully used ultrasonic velocity to determine the maturation time of Mahon cheese. The age can be determined by
\(M(v)=0.0312443 v^{2}-101.39 v+82,264, \quad v \geq 1620\),
where M(v) is the estimated age of the cheese (in days) for a velocity v (m per second). Source: Journal of Food Science.
a. If Mahon cheese ripens in 150 days, determine the velocity of the ultrasound that one would expect to measure. (Hint: Set M(v) = 150 and solve for v.)
b. Determine the derivative of this function when v = 1700 m per second and interpret.
ANSWER:Step 1 of 2
Given that the age can be determined by
\(M(v)=0.0312443 v^{2}-101.39 v+82264, \quad v \geq 1620\)
where \(M(v) \vee(\text { mpersecond }) \text {. }\)
Answer for (a): Now set M(v) = 150 and then solve for v.
So we get,
\(0.0312443 v^{2}-101.39 v+82264=150\)
\(\Longrightarrow 0.0312443 v^{2}-101.39 v+82114=0\)
Now we solve this equation using the quadratic formula.
Therefore,
\(\begin{array}{l}
v=\frac{101.39 \pm \sqrt{(-101.39)^{2}-4(0.0312443)(82114)}}{2(0.0312443)} \\
=\frac{101.39 \pm \sqrt{17.55}}{2(0.0312443)}
\end{array}\)
So, \(v \approx 1690\) meters per second or v = 1555 meters per second.
Since the function is defined only for \(v \geq 1620\), the only solution is 1690 meters per second.
Hence, the velocity of the ultrasound is 1690 meters per second.