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Calculus Word Problem?

Calculus Word Problem? Topic: Solving a distance word problem
June 25, 2019 / By Loyd
Question: A small resort is situated on an island off a part of the coast of Mexico that has a perfectly straight north-south shoreline. The point P on the shoreline that is closest to the island is exactly 4 miles from the island. Ten miles south of P is the closest source of fresh water to the island. A pipeline is to be built from the island to the source of fresh water by laying pipe underwater in a straight line from the island to a point Q on the shoreline between P and the water source, and then laying pipe on land along the shoreline from Q to the source. It costs 2.2 times as much money to lay pipe in the water as it does on land. How far south of P should Q be located in order to minimize the total construction costs? Hint: You can do this problem by assuming that it costs one dollar per mile to lay pipe on land, and 2.2 dollars per mile to lay pipe in the water. You then need to minimize the cost over the interval [0,10] of the possible distances from P to Q.
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Best Answers: Calculus Word Problem?

Jeb Jeb | 7 days ago
If "x" is the distance from point Q to the water source, then the cost is: cost(x) = 1*x + 2.2*sqrt(4^2 + (10-x)^2) cost(x) = x + 2.2*(4^2 + (10-x)^2)^(1/2) cost(x) = x + 2.2*(16 + 100 - 20x + x^2)^(1/2) cost(x) = x + 2.2*(x^2 - 20x + 116)^(1/2) To find the min/max, take the derivative using chain rule: cost'(x) = 1 + 2.2*(1/2)*(x^2 - 20x + 116)^(-1/2) * (2x - 20) cost'(x) = 1 + 1.1*(2x-20) / (x^2 - 20x + 116)^(1/2) cost'(x) = 1 + (2.2x - 22) / (x^2 - 20x + 116)^(1/2) Then solve for the zeros: 0 = 1 + (2.2x - 22) / (x^2 - 20x + 116)^(1/2) -1 = (2.2x - 22) / (x^2 - 20x + 116)^(1/2) -(x^2 - 20x + 116)^(1/2) = (2.2x - 22) (x^2 - 20x + 116)^(1/2) = 22 - 2.2x x^2 - 20x + 116 = (22 - 2.2x)^2 x^2 - 20x + 116 = 22^2 - 2*22*2.2x + (2.2x)^2 x^2 - 20x + 116 = 484 - 96.8x + 4.84x^2 0 = 3.84x^2 - 76.8x + 368 x = (-(-76.8) +/- sqrt(76.8^2 - 4*3.84*368)) / (2*3.84) x = (76.8 +/- sqrt(5898.24 - 5652.48)) / 7.68 x = (76.8 +/- 15.676734354) / 7.68 Thus: x = (76.8 + 15.676734354) / 7.68 = 12.04124145234 OR x = (76.8 - 15.676734354) / 7.68 = 7.958758547656 Since it doesn't make sense to be more than 10 miles from the source, the correct answser is to direct the underwater pipe to a location which is 7.96 miles north of the water source. Alternatively stated, it should be 10-7.96=2.04 miles south of point P .
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Jeb Originally Answered: Calculus word problem help?
So first we have to identify what the problem gives us. We are given a triangle, with a height (h)of two miles, this height is constant. However, the length (l) of the triangle has a constant rate of change of 490, meaning the hypotenuse (c) of the triangle is constantly changing as a function time. Therefore we must relate the side, (l) to the hypotenuse (c) using the Pythagorean theorem. c^2 = l^2+h^2 Now let us plug in what we know to the equation: c^2 = l^2 + 4 Even though we know 'l' is 10, we cannot plug it in yet because the value 10 is not constant as a function like h is. take the derivative of htis equation and we get 2c(dc/dt)=2l(dl/dt)+0 there are four unknown values in this equation, c, (dc/dt), (dl/dt), and l The problem tells us that our instantaneous 'l' value must is 10, and that the rate of change l 'dl/dt' is 490. 'c' can be calculated by using the Pythagorean theorem, to get the sqrt(104). This leaves us with only one variable, dc/dt, which is the rate of change of the hypotenuse, or in otherwords, the distance to the station. dc/dt = 2(10)(490)/(2)/(sqrt(104)) = 480.48mi/h I believe this is correct, I'm in calculus too and this is how I remember solving related rates, and this matches the methods I've seen online, but still... I'm not putting a 100% guarantee on this answer.
Jeb Originally Answered: Calculus word problem help?
The visual image/sketch is this: A right triangle with the rt. angle above the radar site and the horizontal (at 2 miles altitude is represented by x). The Hypotenuse from radar-site to plane "obeys" this condition at ALL TIMES: X² = H² - 2² (Pythag theorem) From this ONE function we seek dH/dt at a certain time. Implicit differentiating... 2x(dx/dt) = 2H(dH/dt) and solving for dH/dt we get (before reducing) [4√6/10] [490] ≈ 480.1mph. = = = = The determination of x at the time the plane is 10 miles from the radar-site should be readily accessible to you; after all we only have ONE function.

Gamliel Gamliel
enable x = distance of base of ladder from domicile. enable y = top of ideal of ladder. x² + y² = 29² = 841 while base of ladder is 21 feet from domicile: y² = 841 - 21² y² = 4 hundred y = 20 x² + y² = 841 Differentiate the two factors with appreciate to t 2x dx/dt + 2y dy/dt = 0 y dy/dt = -x dx/dt while base is 21 feet from the domicile, x = 21, y = 20 Base of ladder is pulled far flung from the domicile at a fee of four feet/sec: dx/dt = 4 y dy/dt = -x dx/dt 20 dy/dt = -21 * 4 dy/dt = -80 4/20 dy/dt = -4.2 So ideal of ladder is shifting down the wall at a fee of four.2 feet/sec -------------------- notice: dy/dt = -4.2 The unfavorable sign shows that ideal of ladder is shifting in downward path. So we are saying that ladder is shifting at a fee of -4.2 feet/sec OR ladder is shifting DOWN at a fee of four.2/sec
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