[URGENT] Maxima and Minima Help (1 Viewer)

[cherii2007]

I have my exam tomorrow (Geometrical Applications of Calculus). These are questions out of the Maths in Focus textbook...Exercise 1.8. Q 11, 13, 14.

11. A silo in the shape of a cylinder is required to hold 8600 m&3 of wheat.
a) Find an equation for the surface area of the silo in terms of the base radius.
b) Find the minimum surface area required to hold this amount of wheat, to the nearest square metre.

13. A poster consists of a photography bordered by a 5 cm margin. The area of the poster is to be 400 cm^2.

a) Show that the area of the photograph is given by the equation A = 500-10x-4000/x I actually know how to do this, but I typed it out for wholeness.
b) Find the maximum area possible for the photograph.

14. The sum of the dimensions of a box with a sequare base is 60 cm. Find the dimensions that will give the box a maximum volume.

ANS/
11. a) S = 2 pi r^2 + 17200/r b) 2323.7 m^2
13. a) Working out, working out. b) 100 cm^2
14. 20 x 20 x 20

I might have more questions in an hour or two. Thanks!

 

[Luxxey]

11. A silo in the shape of a cylinder is required to hold 8600 m&3 of wheat.
a) Find an equation for the surface area of the silo in terms of the base radius.
b) Find the minimum surface area required to hold this amount of wheat, to the nearest square metre.

We know that A = πr^2 x h = 8600 m^3
Rearranging for h, h = 8600/πr^2

Surface area equation is given by S = 2πr^2 + 2πrh
(as in, two circles and one rectangle whose width is equal to the circumference of a circle)

Substituting the value of h in terms of the base radius,
S = 2πr^2 + 2πr(8600/πr^2)
= 2πr^2 + 17200/r


Differentiating,
dS/dr = 4πr - 17200/r^2

Making it equal to zero to find stationary points,
4πr = 17200/r^2
4πr^3 = 17200
r^3 = 17200/4π
r = cube root (17200/4π) = 11.102979...

Differentiating for the second derivative,
d2S/dr2 = 4π + 24400/r^3

Making it equal to zero,
4πr^3 = -34400
r^3 = -34400/4πr

As 4πr is always positive, therefore r^3 is always negative, meaning r is always negative. Hence r = 11.102979... is a minimum.

Subbing back into your S equation for minimum surface area,
S = 2323.7 m^2 (1 d.p.)



Sorry but that's all I had time to quickly do! Good luck.

 

[fullonoob]

We know that A = πr^2 x h = 8600 m^3
Rearranging for h, h = 8600/πr^2

Surface area equation is given by S = 2πr^2 + 2πrh
(as in, two circles and one rectangle whose width is equal to the circumference of a circle)

Substituting the value of h in terms of the base radius,
S = 2πr^2 + 2πr(8600/πr^2)
= 2πr^2 + 17200/r


Differentiating,
dS/dr = 4πr - 17200/r^2

Making it equal to zero to find stationary points,
4πr = 17200/r^2
4πr^3 = 17200
r^3 = 17200/4π
r = cube root (17200/4π) = 11.102979...

Differentiating for the second derivative,
d2S/dr2 = 4π + 24400/r^3

Making it equal to zero,
4πr^3 = -34400
r^3 = -34400/4πr

As 4πr is always positive, therefore r^3 is always negative, meaning r is always negative. Hence r = 11.102979... is a minimum.

Subbing back into your S equation for minimum surface area,
S = 2323.7 m^2 (1 d.p.)



Sorry but that's all I had time to quickly do! Good luck.

underlined your minor typo an alternative method is after you get r, sub it into the second derivative of the equation. If the second derivative is > 0 it is min, <0 = max.

 

[marmsie]

13 b)

A = 500 - 10x - 4000 / x
= 500 - 10x - 4000x^-1

A' = -10 + 4000x^-2

Set A' = 0 for turning points

0 = -10 + 4000x^-2

1 / 400 = x^-2

x^2 = 400

Therefore, x = 20 and x = -20 (but we can ignore the -ve answer as that is not a physically possible result in this case).

To show x = 20 gives a maximum area:

A'' = -8000x^-3

x = 20:
A'' = -1 < 0

Therefore x = 20 gives a maximum area.

So the maximum area is given by:

A = 500 - 10 * 20 - 4000 / 20 = 100 cm^2


14)
Let the dimensions of the box be:
Length = Depth = x
Height = h

We know from the question that the sum of the dimensions equals 60, which gives us a way of eliminating one of the variables later on.

2x + h = 60
h = 60 - 2x

Volume of the box is given by:

V = x^2 * h
= x^2 * (60 - 2x)
= 60x^2 - 2x^3

Going through the usual process of finding maximum and minimum points:

V' = 120x - 6x^2

0 = 120x - 6x^2
0 = 20x - x^2
0 = x (20 - x)

Therefore turning points are x = 0, x = 20

V'' = 120 - 12x

x = 0:
V'' = 120 > 0 (Therefore x = 0 gives a minimum volume, which is obvious anyway)

x = 20:
V'' = -120 < 0 (Therefore x = 20 gives a maximum volume)

Therefore the dimensions that give a maximum volume are

20 x 20 x (60 - 2*20) = 20 x 20 x 20

Good luck in your test.

 

[cherii2007]

haha they seem so simple now that there's actual working out for it. Thanks, Luxxey, fullonoob and marmsie. Fingers crossed I do okay. This probably won't be the last you've seen of me. (x It's a bit late now, so I'll just ask questions to my maths teacher tomorrow...