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Math Help - Difficulty with Optimisation (Maxima and Minima) and Integration Problems

  1. #1
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    Difficulty with Optimisation (Maxima and Minima) and Integration Problems

    Hi there,

    There are a few optimisation and integration problems I can't seem to solve. Please help me, exam is on Tuesday... (Don't have to solve it, just tell me how to solve the questions )

    1. P is a moveable point on the line y=7-x. Find the coordinates of P when it is closest to the point (2,4).
    2. A small boat moving at x km/h uses fuel at a rate that is approximated by the function q=8+x^2/50, where q is measured in litres/hour. Determine the speech of the boat for which the amount of fuel used for any given journey is least.
    3. An 8m high fence stands 3m from a large vertical wall. Find the length of the shortest ladder that will reach over the fence to the wall behind, as shown in the diagram.Difficulty with Optimisation (Maxima and Minima) and Integration Problems-1.png
    4. A motorbike is 30km directly west of a car and begins moving east at 90km/h. At the same instant, the car begins moving north at 60km/h. What will be the minimum distance between the vehicles.

    This is the integration question:
    5. The cross-section of a channel is parabolic. It is 3 metres wide at the top and 2 metres deep. Find the depth of water, to the nearest cm, when the channel is half full. (For this one, I got that when the channel is half-full, 2 squared metres of the cross-section need to be covered, but I can seem to solve the question. Do I have to integrate with respect to y?)

    Thank you so much for your help
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  2. #2
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    Re: Difficulty with Optimisation (Maxima and Minima) and Integration Problems

    I will help with first two problems, but before this here is the link to get a better touch of optimization problems.
    1.P is a moveable point on the line y=7-x. Find the coordinates of P when it is closest to the point (2,4).
    Since P is a point on line y=7-x, then coordinates of point P are (x,7-x).
    Now, calculate distance between point P and point (2,4):
    d=\sqrt{(x-2)^2+(7-x-4)^2}=\sqrt{x^2-4x+4+9-6x+x^2}=\sqrt{2x^2-10x+13}
    Clearly if we find minimum of d then it will also be minimum for d^2. Therefore, for simplicity we will find minimum of f(x)=d^2=2x^2-10x+13
    Find derivative: f'(x)=4x-10
    Set f'(x)=0: 4x-10=0 or x=2.5.
    Then y=7-x=7-2.5=4.5

    Therefore, the closest point is P=(2.5,4.5). d=\sqrt{(2.5-2)^2+(4.5-4)^2}=0.5\sqrt{2}

    2.A small boat moving at x km/h uses fuel at a rate that is approximated by the function q=8+x^2/50, where q is measured in litres/hour. Determine the speech of the boat for which the amount of fuel used for any given journey is least.

    This means that if boat is moving at x km/h then it uses f(x)= \frac{q}{x}=\frac8x+\frac{x}{50} litres per kilometer.
    We must minimize this quantity: f'(x)=-\frac{8}{x^2}+\frac{1}{50}
    Set f'(x)=0: -\frac{8}{x^2}+\frac{1}{50}=0 or x^2=400 or x=20 and x=-20. x=-20 is impossible because speed cannot be negative.
    Therefore, only x=20 is applicable. Now we need to check that x=20 is indeed minimum.
    For this we will use second derivative test: since f''(x)=\frac{16}{x^3} then f''(20)=\frac{16}{8000}>0 therefore x=20 is indeed minimum.
    Answer: x=20 kilometers per hour.
    Thanks from chessweicong
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  3. #3
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    Re: Difficulty with Optimisation (Maxima and Minima) and Integration Problems

    Quote Originally Posted by chessweicong View Post
    Hi there,

    There are a few optimisation and integration problems I can't seem to solve. Please help me, exam is on Tuesday... (Don't have to solve it, just tell me how to solve the questions )

    1. P is a moveable point on the line y=7-x. Find the coordinates of P when it is closest to the point (2,4).
    2. A small boat moving at x km/h uses fuel at a rate that is approximated by the function q=8+x^2/50, where q is measured in litres/hour. Determine the speech of the boat for which the amount of fuel used for any given journey is least.
    3. An 8m high fence stands 3m from a large vertical wall. Find the length of the shortest ladder that will reach over the fence to the wall behind, as shown in the diagram.Click image for larger version. 

Name:	1.png 
Views:	2 
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ID:	24052
    4. A motorbike is 30km directly west of a car and begins moving east at 90km/h. At the same instant, the car begins moving north at 60km/h. What will be the minimum distance between the vehicles.

    This is the integration question:
    5. The cross-section of a channel is parabolic. It is 3 metres wide at the top and 2 metres deep. Find the depth of water, to the nearest cm, when the channel is half full. (For this one, I got that when the channel is half-full, 2 squared metres of the cross-section need to be covered, but I can seem to solve the question. Do I have to integrate with respect to y?)

    Thank you so much for your help
    For question 1 let the point be (a,b) It will be closest to (2,4) when the line joining (a,b) to (2,4) is perpendicular to y=7-x
    which has gradient -1 So we require a gradient of +1 So (b-4)/(a-2)=1 Also (a,b) is on y=7-x So b=7-a Solve these two equations
    Thanks from chessweicong
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  4. #4
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    Re: Difficulty with Optimisation (Maxima and Minima) and Integration Problems

    Thank you guys so much Is there anyone who knows how to do the other ones? I tried but they are all too difficult for me
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  5. #5
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    Re: Difficulty with Optimisation (Maxima and Minima) and Integration Problems

    3. Let x be distance between bottom of ladder and fence. Let y be the difference between height of fence and height of top of ladder. Let L be the length of ladder.
    Then by Pythagorean theorem: L^2=(x+3)^2+(y+8)^2
    From similar triangles \frac{x}{8}=\frac{x+3}{y+8}
    From this y+8=\frac{8(x+3)}{x}
    So, L^2=(x+3)^2+(\frac{8(x+3)}{x})^2
    Or L^2=(x+3)^2+(8+\frac{24}{x})^2
    Note, that L and L^2 share same minimum.
    So, we will minimize f(x)=L^2:
    f'(x)=2(x+3)+2(8+\frac{24}{x})(\frac{-24}{x^2})
    I think next steps are clear.
    4. Let's take origin to lie where car stands and let t be the time.
    Then position of car at any time t is 60t and position of motorbike is 90t-30.
    Distance between the by pythagorean theorem is d^2=(60t)^2+(90t-30)^2
    Again we minimize f(t)=d^2

    f'(t)=2*60t*60+2(90t-30)*90
    Set f'(t)=0: 2*60t*60+2(90t-30)*90=0 or t=\frac{3}{13}
    Minimum distance is d=\sqrt{(60t)^2+(90t-30)^2}=\sqrt{(60*\frac{3}{13})^2+(90*\frac{3}{13}-30)^2}=\frac{60}{\sqrt{13}} kilometers
    Thanks from chessweicong
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  6. #6
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    Re: Difficulty with Optimisation (Maxima and Minima) and Integration Problems

    Thank you so much!
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