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The illumination of an object by a light source is directly proportional to the strength of the source and inversely proportional to the square of the distance from the source. If two light sources, one three times as strong as the other, are placed $ 10 ft $ apart, where should an object be placed on the line between the sources so as to receive the least illumination?

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03:25

Wen Zheng

01:14

Amrita Bhasin

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 7

Optimization Problems

Derivatives

Differentiation

Volume

Baylor University

University of Michigan - Ann Arbor

University of Nottingham

Idaho State University

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All right, We've got a question here. Describe the illumination of an object by a light source directly proportional to the strength of the source inversely proportional to the square of the distance from the source. Excuse me. There's two light sources call this light source. One nice was to that are 10 ft apart. Then where should we place no object on the line between the sources so that they received the least illumination? Eso You know that one of them is three times stronger than the other one is weak. Hypothetically, say that l is a strong one. That would be the L one would be equal to 32 And here we can write out any an equation or the intensity of the light. Of course, you basically we wanna place objects somewhere. Yeah. Yeah. You want to place this object at some point and calculate. So if we wrote out an equation for the intensity of light, we know that the intensity of the light has given us the strength. They were told that it was fortunate to strength a distance squared. Uh huh. This year, we call x issue because 10 minus X right now. we will start filling in strength would have three l mhm. Mhm. Oh, two. Okay, over the distance. Where? Which in this case would be X. Where? Mhm. Uh huh. Well, to over. Mhm. Mhm. Yeah. Turn minus X squared. Thanks. Without the strength over the distance. Where? Strength over the distance. Where? Mhm. That is our intensity of light equation. Mhm. And now if we go ahead and take the derivative intensity light, you will get mhm 32 negative too Backs to another three waas to I was negative Two. Okay, times 10 minus x two in every three. Mhm! Mhm! All right. Equals to zero. Mhm. Mhm. Yeah, e Here. This should be You have six out to over. Excellent. Third Waas. Excuse me. Minus negative to to 10 months. Cute zero. Mhm. All right, now let's move. Theo l is on one side and the excess on the other side. We could do how to execute you Go to positive Mhm. Okay, So negative. Negative should be a positive. This is a negative. Negative. Become lost negative to our to 10 minus to be huge or you divide everything by negative too. Three l two. What you hell to over until three. Tend to the X himself. Some reason im writing through here. You should be an X. Yeah, mhm. We can divide everything by two here. So you'll have three over Execute. It was one over and one x three. Another three. Speak well to one over. Execute 10 minus X to the okay. Mm hmm. Mhm. Mhm. Mhm. Mhm. Sure. Let's go and divide by three. So x cubed. Mhm. Yeah. Yeah. We multiply X cubed on both sides and then multiplied 10 to the 3rd 10 minus X to the third on both sides. Don't get execute here and be multiplied by 10 miles X. All right, If we saw her X, we would get the square root. So cue route off. Three. 10 minus. Excellent. Who? Just the same thing as you wrote three multiplied by 10 minus x mhm. We're good. 10 times Cube root three minus cube root three x. And then we can add the access together. So you know what? You boot Really same thing. That's 1.44 plus one to or too x. It's equal to turn three. X is equal to turn Huber three over to 442 Work that into my handy dandy calculator. We'll get 5.91 Yeah, that'll be our distance, x mhm. And then our distance here would be 10 minus that which is 4.9 five point nine one. All right, well, I hope that clarifies the question. Thank you so much for watching.

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