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For a fish swimming at a speed $ v $ relative to the water, the energy expenditure per unit time is proportional to $ v^3 $. It is believed that migrating fish try to minimize the total energy required to swim a fixed distance. If the fish are swimming against a current $ \mu(\mu < v) $, then the time required to swim a distance $ L $ is $ L/(v - \mu) $ and the total energy $ E $ required to swim the distance is given by$$ E(v) = av^3 \cdot \dfrac{L}{v - \mu} $$where $ a $ is the proportionality constant.(a) Determine the value of $ v $ that minimizes $ E $.(b) Sketch the graph of $ E $.Note: This result has been verified experimentally; migrating fish swim against a current at a speed of 50% greater than the current speed.

a) $v=\frac{3}{2} u$b) $\left(\frac{3}{2} u, E\left(\frac{3}{2} u\right)\right)$

04:08

Wen Z.

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 7

Optimization Problems

Derivatives

Differentiation

Volume

Missouri State University

Oregon State University

University of Nottingham

Boston College

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in this problem we have to use our knowledge of differentiation Thio optimize a specific function were given and the function that were given deals with various things like distance and, um speed. So we're given this function will call Eve equals a V cubed times l over v minus you. Well, we can simplify that a little bit to say e of e equals a V cubed times l Oliver v minus you. What we want is we want the derivative equal to zero. So we'll take the derivative of this function. We will set it equal Thio zero we'll have to a V cubed times l minus three a u L V squared equal to zero. We could simplify a little bit and cancel out some terms will have to v cubed minus three u V squared equal to zero and then we can factor out a common a common factor. Essentially, we have the square and will multiply that by the quantity to V minus three You equal to zero and then we can solve for V pretty easily. These three over to you And now for be, we're told. Well, let's find this information by using a graph. We need that coordinate point of where? Yeah, our values of X and Y satisfy the equation. So this is a little bit tricky, really. You can test different values and see what you get. These values work well. We have l equal 100. You equal five and a equal 50.1 And this is the graph that we get. So this coordinate point is essentially the answer to part B. This satisfies the specific speeds and the derivative, Um, that we need. So I hope that this problem helped you understand a little bit more about optimization and how we can go about optimizing a function using differentiation and its associated graph.

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