Maxima And Minima (1 Viewer)

[Lukybear]

6. The cost per hour of running a pleasure cruiser is $(V^2/40 + 10)
a) For a trip of D nautical miles show that the running cost is $(D/V(V^2/40 + 10))

b) What is the most economical speed for running the cruiser on this trip.

Answer: 20

I have no idea. Help?

 

[Lukybear]

Also;

11. A window consists of a rectangle surmounted by a semi-circle having as its diameter the width of the rectangle. If the perimeter of the window is t metres, find the greatest possible area of the window.

 

[blaze.bluetane]

A.

The cost per hour of running a pleasure cruiser is $(V^2/40 + 10) so when the cost per hour is multiplied by the number of hours, we are left with the total cost
(V=velocity must be measured in miles per hour)

C=(V^2/40 + 10)t
where t is the time taken in hours (C=cost per hour)

Velocity is the displacement in a certain time frame,
Mathematically, V=D/t
and therefore t=D/V

Substituting in,
A trip of D nautical miles will take D/V hours
The cost of this trip will be C=[D/V(V^2/40+10)]


B.

C=[D/V(V^2/40+10)]

dC/dV=(1/40-10/V^2)D

d^2C/dV^2=(20/V^3)D
We cannot undo our travel so V>0 and consequently d^2C/dV^2>0
The second derivative is concave down for the domain concerned
Therefore minimum cost at d^2C/dV^2=0

(1/40-10/V^2)D=0
V^2=400
V>0 so V=20

 

[Lukybear]

Thxs blazebluetane. Now Ive realised that D is negligible. Couldnt see it before.

 

[benv]

A.

The cost per hour of running a pleasure cruiser is $(V^2/40 + 10) so when the cost per hour is multiplied by the number of hours, we are left with the total cost
(V=velocity must be measured in miles per hour)

C=(V^2/40 + 10)t
where t is the time taken in hours (C=cost per hour)

Velocity is the displacement in a certain time frame,
Mathematically, V=D/t
and therefore t=D/V

Substituting in,
A trip of D nautical miles will take D/V hours
The cost of this trip will be C=[D/V(V^2/40+10)]


B.

C=[D/V(V^2/40+10)]

dC/dV=(1/40-10/V^2)D

d^2C/dV^2=(20/V^3)D
We cannot undo our travel so V>0 and consequently d^2C/dV^2>0
The second derivative is concave down for the domain concerned
Therefore minimum cost at d^2C/dV^2=0

(1/40-10/V^2)D=0
V^2=400
V>0 so V=20

Nice Gurll =]

 

[em_angel23]

A.

B.

C=[D/V(V^2/40+10)]

dC/dV=(1/40-10/V^2)D

d^2C/dV^2=(20/V^3)D
We cannot undo our travel so V>0 and consequently d^2C/dV^2>0
The second derivative is concave down for the domain concerned
Therefore minimum cost at d^2C/dV^2=0

(1/40-10/V^2)D=0
V^2=400
V>0 so V=20

sorry, i'm having trouble differentiating, what rule are you using?
thanks in advance!

 

[Lukybear]

consider the c(x) as d(v/40 + 10v^-1) through expanding and then differentiate.

Btw: its tricky cause there is two variable, i.e. d and v, but d can be considered as a constant.

 

[em_angel23]

Thanks!

 

[Lukybear]

One more question. INteressted in how you guys would set it out.

Find the largest rectangle that can be inscribed in the region bounded by the parabola with equation y = 4-x^2 and the x-axis. This is done purely using differenttiation.

 

[Drongoski]

The parabola is upside-down of y=x² pushed up 4 units so that vertex is (0,4)
Since rectangle is symmetric about the y-axiset the right base corner of rectangle be (x,0) so that the other base corner is (-x,0).

.: area of rectangle A(x) = 2x . y = 2x(4 - x²) = 8x - 2x³

.: A'(x) = 8 - 6x² = 0 @ x = sqrt(4/3)

and A"(x) = -12x which is negative for x=sqrt(4/3)

.: we have a max A(x) @ x = sqrt(4/3)

.: The largest rectangle correspond to width: 2x = 2sqrt(4/3) and height (4-4/3) = 8/3

and the max area = 2 x sqrt(4/3) x (8/3) = 32sqrt(3)/9 sq units.

Hopefully correct.