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Topic: **Time and work problem solving methods****Question:**
So I have this optimization problem to do. I tried to get some help from school tutors, they did not know how to do it. "Must be solved by a calculus method"
You are at the southernmost point of a circular lake of radius 1 mile. Your plan is to swim a straight course to another point on the shore of the lake, then jog to the northernmost point. You can jog twice as fast as you can swim. Find the route you must take to minimize the time this trip will take and find the time in minutes.
I have never seen an optimization like this and I am having a hard time just to set it up. I need to solve this by a CALCULUS method.
Please, I need some help.

June 20, 2019 / By Devan

Swim to a point on the circle angle 2A (to avoid fractions) round from the southern end. Then jog round the arc (pi - 2A) that remains. (We must work in radians due to differentiation of trig function.) A diagram will show that the straight part = 2*cosA. The time for the whole exercise is T = 2*cosA + (1/2)*(pi - 2A) = 2*cosA + pi/2 - A (The 1/2 at the front of the second term is due to speed being twice as fast.) dT/dA = -2*sinA - 1 For minimum we need dT/dA = 0 ----> sinA = -1/2 ----> A = -30 degrees. 2A = -60 degrees or 60 degrees round the other way. This is where you swim to and then jog the rest.

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The distance that you will swim will be the third leg of an isosceles triangle. Splitting this triangle down the middle gives two congruent triangles with hypotenuse of 1 mile. Let's call the base of each of these triangles 'x'. When you draw it, you'll notice that: x/1 = cosΘ ==> 2x = 2cosΘ. Therefore, the distance that you will swim is 2cosΘ, since there are two triangles. The rate that you'll travel this distance is 'r'. This distance that you'll jog is pi minus the arc-length subtended by the angle pi - 2Θ. (pi - (pi - 2Θ)) = 2Θ The rate that you'll jog this distance is 2r. Now let's use: time = distance / rate. t = 2cosΘ/r + 2Θ/2r. Differentiating yields: t ' = -2sinΘ/r + 1/r. Now let's set our numerator equal to 0. -2sinΘ + 1 = 0 sinΘ = 1/2 Θ = pi/6 t(pi/6) = 2cos(pi/6)/r + 2(pi/6)/2r t(pi/6) = (sqrt3)/r + pi/6r. Since you didn't give the rate, I can't find the exact time in minutes. But notice that the rate at which he swims must be slow. Let's say that it is 1/10 mi/min. Then we'd have: t(pi/6) = (sqrt3)/(1/10) + pi(6/10) t(pi/6) = 17.3 + 5.23 = 22.5. Therefore, the time is reasonable. Essentially, you'll swim a distance of sqrt3 miles to the point on the circle that makes an angle of 2pi/3 with the point that you started from. Then you'll jog a distance of pi/3 around the edge of the lake. Note that the straight-line distance is 2 miles, since radius is 1. The distance that you'll travel, however is pi/3 + sqrt3 = 2.77 miles, which makes sense. Hope this helped.

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y-10 = m(x-6) is the equation of all strains via (6,10) the are reduce off is a triangle to grasp the peak and base of this triangle we need to calculate the intersections with x-axis and y-axis. x=zero => y=10-6m y=zero => x=6-10/m so the field is (10-6m)(6-10/m)/two = (60+60-36m-one hundred/m)/two = 60 - 18m - 50/m We need to take the by-product and placed it 0 to discover a minimal = -18 + 50/m² = zero => 50/m² = 18 => m² = 50/18 = 25/nine => m = +- five/three to look if it is a minimal, derive once more -one hundred/m³ > zero if m

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You find the extrema of the function as you normally would: take the derivative with respect to I, set the derivative equal to zero, solve for the values of I that make that equation true, then check to see if those extrema are maxima or minima. P(I) = (100*I)/(I^2 + I + 4) dP(I)/dI = 100/((I^2 + I + 4) - (100*I)*(2*I + 1)/(I^2 + I + 4)^2 dP/dI = 100*(4 - I^2)/(I^2 + I + 4)^2 Set this equal to zero and solve for I: 0 = 100*(4 - I^2)/(I^2 + I + 4)^2 0 = 4 - I^2 I = +2, I = -2 Because I is the light intensity, the negative root has no physical meaning. Now check to see if this is a maximum. The easiest way is simply to plot the P(I) vs I, but you can also take the second derivative and determine its sign at I = 2. If the second derivative is negative, the extremum at I=2 is a maximum. If it's positive, then the extremum is a minimum. P'' = 200*(I^3 - 12I - 4)/(I^2 + I + 4)^3 P''(2) = -4 so this is, in fact, a maximum. (The extremum at I=-2 is a miminum.)

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