[Technology Exercise] Grinding engine cylinders Before contrac...

Question

[Technology Exercise] Grinding engine cylinders Before contracting to grind engine cylinders to a cross-sectional area of 9in², you need to know how much deviation from the ideal cylinder diameter of c = 3.385in. you can allow and still have the area come within 0.01in² of the required 9in². To find out, you let A=π(x/2)² and look for the largest interval in which you must hold x to make |A − 9| ≤ 0.01. What interval do you find?

Accepted Answer

Welcome back, everyone. In this problem, a homeowner decides to construct a round swimming pool with a cross-sectional area of 50 square meters. To comply with local regulations, the deviation from the ideal pool diameter of C equals 7.98 m must be within 0.08 square meters of the planned area. Using the formula A equals pi multiplied by a half of X squared, where X is the diameter. Identify the largest interval where X must be kept to make the absolute value of A minus 50 less than or equal to 0.08. Round your answer to three decimal places. A says the interval is open 7.970, 7.985 closed parentheses. B says it's from 7.973 to 7.987. C from 7.972 to 7.985, and D from 7.978 to 7.986. Now what are we trying to figure out here? Well, we're trying to identify the largest interval, OK, around the point where C equals 7.98 m, where the area satisfies this inequality. The absolute value of A minus 50 is less than R equal to 0.08. Now fortunately in our problem we are given a formula for the area. So if we substitute that area of formula into the expression, then really we're looking for the absolute value of pi multiplied by x divided by 2 squad minus 50. To be less than or equal to 0.08. So let's try to simplify this, uh, expand and find and accept to be error range, and then solve for the possible values of X for or boundaries. Now, if we think about it here, if we expand our term pi multiplied by a half of X2, OK, it's gonna become pi X2 divided by 4 minus 50. That's the absolute value, less than or equal to 0.08. And know if we, if we look at our bounds here. Or if we write this in terms of bounds, this means the acceptable AR range is going to be between 49.92, OK. And 50.08. OK? In other words, 50 plus R minus 0.08. Now, if we're gonna solve for our values of X here, let's take both bounds. First, let's start with our lower bound, OK? Now for our lower bound, this means then that that is where a, let me write this in another way. Where A equals 49.92 square meters, OK. Then We're gonna have pi X square divided by 4 to be equal to 49.92. Which means that if we multiply both sides by 4 divided by pi, X squared equals 199.68 divided by pi, which means that X is gonna be equal to the root of 199.68 divided by pi and no from this lower bound, OK. Then We can get that to be approximately equal to 7.972 m to 4 significant figures and note that we write it to 4 significant figures or round it up to 3 decimal places to keep the area within the error tolerance of 0.08 square meters. Now, if we do on the other hand and take the upper bound, OK, let's take the upper bound here, then we can figure out the value of X when A equals 0, sorry, when A equals 50.08. No, in this case, then again we say pi x 2 divided by 4 E equals 50.08, which means that X2 equals 50.08 multiplied by 4. right that is 200.32 divided by pi. Which means X is gonna be equal to the root of 200.32 divided by pi, which is approximately equal to 7.985. What is it, what uh meters, my apologies. Thus, this tells us then that our interval, OK. The required interval. Is between 7.972 m. 2, let's just write it the same way that the answers were answers were to 7.985 m. If we look back at our answer choices, that tells us then that C is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Area of a Circle

Deviation and Tolerance

Inequalities in Calculus

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