Differentiation (1 Viewer)

[FDownes]

I'm having a bit of trouble with the second half of this question...

And open rectangular box has four sides and a base, but no lid. The box has a base of dimensions 3x cm and 2x cm, and a height of y cm.

a) Write dow formulae for the outer surface area A cm² of the box and the volume V cm³.
= A = 6x² + 10xy
= V = 6x²y

b) It is known that A = 240. Eliminate y to obtain a formula V(x) for the volume as a function of x.
= V = [18x(40 - x²)]/5

From here onwards I'm having trouble...

c) Show that x = 2√10

d) Find the value of x for which V is a maximum and verify the maximum value is 64√30

 

[3unitz]

c) Show that x = 2√10

sub it into the equation for V and it should equal 0 (i.e. satistfies the equation).

d) Find the value of x for which V is a maximum and verify the maximum value is 64√30

V ' (x) = 144 - (54/5)x^2
0 = 144 - (54/5)x^2
x = 12 /√(54/5)
x = 2√30/3

check second derivative for maximum, then sub it into V

V_max = (18/5) (2√30/3) (40 - (2√30/3)^2)
= (12√30/5) (40 - 40/3)
= (12√30/5) (80/3)
= 64√30

 

[FDownes]

Thanks for that, now time for another...

A cylinder of radius r cm and height h cm is inscribed in a cone with base radius 3 cm and height 10 cm, as in the diagram.

a) Show that the volume V of the cylinderis given by V = [10(pi)r²(3 - r)]/3

b) Hence find the values of r and h for the cylinder which has maximum volume.

c) What is the maximum volume?

 

[3unitz]

a) Show that the volume V of the cylinderis given by V = [10(pi)r²(3 - r)]/3

here we just need to get h in terms of r.

View attachment 16709

at x = r, h = -10r/3 + 10

.'. V = pi r^2 (-10r/3 + 10)

= (pi)r^2(30 - 10r)/3

= 10(pi)r^2(3 - r)/3

b) Hence find the values of r and h for the cylinder which has maximum volume.

V = 10(pi)r^2 - 10(pi)r^3/3

dV/dr = 20(pi)r - 10(pi)r^2

0 = 20(pi)r - 10(pi)r^2

0 = r(2 - r)

r = 2 (check for max V''(2) < 0)

h = -10(2)/3 + 10

h = 10/3

c) What is the maximum volume?

V = 10(pi)r^2 - 10(pi)r^3/3

sub r = 2,

V_max = 10(pi)2^2 - 10(pi)2^3/3

V_max = 40pi/3 units³

 

[FDownes]

Thanks again, here's another;

A plane is to fly 3000km at a constant speed of v km/h. When flying at v km/h the plane consumes fuel at the rate of (50 + 10^-6v³) litres per hour.

a) Show that on a journey of 3000km at a speed of v km per hour, the expression for the total amount of fuel used, A litres, is given by A = 150000/v + (3v²)/1000.

b) Find the speed for the greatest fuel economy and the amount of fuel used at this speed. Give both answers correct to 3 significant figures.

 

[lolokay]

number of hours = distance/velocity = 3000/v
.'.fueld used = (50 + 10-6v3)*(3000/v)
= 150000/v + (3v^2)/1000

b) A' = -1.5*10^5 v^-2 + 6*10^-3 v = 0
v^3 = 2.5*10^7

This is a minimum for A as A->infinite as v->0 or infinite
Then just calculate values for v and A.

 

[FDownes]

Thanks. And now, time for another;

A piece of wire of length 5 meters is bent to form the hypotenuse and side of a right angled triangle ABC. Let the length of the side AB be x meters. What is the maximum possible area of the triangle?

What I know so far;

The hypotenuse (AC) = 5 - x
BC = √(25 - 10x)
Area of the triangle = (x/2)√(25 - 10x)

My problem is in differentiating the area. By my working it should be (-10x)/[4√(25 - 10x)], but I'm not sure if that's correct.

 

[lyounamu]

Thanks. And now, time for another;

A piece of wire of length 5 meters is bent to form the hypotenuse and side of a right angled triangle ABC. Let the length of the side AB be x meters. What is the maximum possible area of the triangle?

What I know so far;

The hypotenuse (AC) = 5 - x
BC = √(25 - 10x)
Area of the triangle = (x/2)√(25 - 10x)

My problem is in differentiating the area. By my working it should be (-10x)/[4√(25 - 10x)], but I'm not sure if that's correct.

Yes your working out is corret.
I will just expand on that:
dA/dx = (x/2)square root(25-10x)
= x/2 . 1/(2square root(25-10x) . -10 + square root(25-10x) . 1/2
The maximum area occurs when dA/dx = 0
i.e. x/2 . 1/(2square root(25-10x) . -10 + square root(25-10x) . 1/2 = 0
-5x/(2square root(25-10x)) + (square root(25-10x))/2
Arranging it will get you: (square root(25-10x))/2 = 5x/(2square root(25-10x))
25-10x = 5x
15x = 25
x = 25/15
= 5/3

Area of the triangle = ((5/3)/2)√(25 - 10 . 5/3) = 25square root(3)/18