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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 :D)

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.Attachment 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 :D

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):

$\displaystyle d=\sqrt{(x-2)^2+(7-x-4)^2}=\sqrt{x^2-4x+4+9-6x+x^2}=\sqrt{2x^2-10x+13}$

Clearly if we find minimum of d then it will also be minimum for $\displaystyle d^2$. Therefore, for simplicity we will find minimum of $\displaystyle f(x)=d^2=2x^2-10x+13$

Find derivative: $\displaystyle f'(x)=4x-10$

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). $\displaystyle d=\sqrt{(2.5-2)^2+(4.5-4)^2}=0.5\sqrt{2}$

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 $\displaystyle f(x)= \frac{q}{x}=\frac8x+\frac{x}{50}$ litres per kilometer.

We must minimize this quantity: $\displaystyle f'(x)=-\frac{8}{x^2}+\frac{1}{50}$

Set f'(x)=0: $\displaystyle -\frac{8}{x^2}+\frac{1}{50}=0$ or $\displaystyle x^2=400$ 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 $\displaystyle f''(x)=\frac{16}{x^3}$ then $\displaystyle f''(20)=\frac{16}{8000}>0$ therefore x=20 is indeed minimum.

Answer: x=20 kilometers per hour.

Re: Difficulty with Optimisation (Maxima and Minima) and Integration Problems

Quote:

Originally Posted by

**chessweicong** 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 :D)

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.

Attachment 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 :D

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

Re: Difficulty with Optimisation (Maxima and Minima) and Integration Problems

Thank you guys so much :D Is there anyone who knows how to do the other ones? I tried but they are all too difficult for me :(

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: $\displaystyle L^2=(x+3)^2+(y+8)^2$

From similar triangles $\displaystyle \frac{x}{8}=\frac{x+3}{y+8}$

From this $\displaystyle y+8=\frac{8(x+3)}{x}$

So, $\displaystyle L^2=(x+3)^2+(\frac{8(x+3)}{x})^2$

Or $\displaystyle L^2=(x+3)^2+(8+\frac{24}{x})^2$

Note, that L and $\displaystyle L^2$ share same minimum.

So, we will minimize $\displaystyle f(x)=L^2$:

$\displaystyle f'(x)=2(x+3)+2(8+\frac{24}{x})(\frac{-24}{x^2})$

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 $\displaystyle d^2=(60t)^2+(90t-30)^2$

Again we minimize $\displaystyle f(t)=d^2$

$\displaystyle f'(t)=2*60t*60+2(90t-30)*90$

Set f'(t)=0: $\displaystyle 2*60t*60+2(90t-30)*90=0$ or $\displaystyle t=\frac{3}{13}$

Minimum distance is $\displaystyle d=\sqrt{(60t)^2+(90t-30)^2}=\sqrt{(60*\frac{3}{13})^2+(90*\frac{3}{13}-30)^2}=\frac{60}{\sqrt{13}}$ kilometers

Re: Difficulty with Optimisation (Maxima and Minima) and Integration Problems