Maths question ~

First, come up with a formula that computes the volume of the box given the length of the sides of the squares. If the length of the sides of the squares is S, what are the three dimensions of the box?

hmm well this is wot i came up with:

if I assigned the length of the square to be x then:

length = 15 - 2x
breadth = 8 -2x
height = x

then generate a formula for the volume = l x b x h

volume = ( 15 - 2x ) ( 8 - 2x) * (x)
gives: 120x-46x^2+ 4x^3

and then u have to differentiate to find the max right? But I can't seem to factorise the differentiated equation when setting to 0. Meaning I can't seem to solve for x. Can help?

but if you make x=2

volume = ( 15 - 2x ) * ( 8 - 2x) * (x)

that gives you 11*4*2=88 wich is the max volume.

so the squares are 2*2.

I haven't checked your earlier work, but your general approach is correct. Differentiating:

V = 120x-46x^2+ 4x^3
dV/dx = 120-92x + 12x^2
dV/dx = 0.

Dividing through by 4 and rearranging:

3x^2 -23x + 30 = 0.

Then, applying the quadratic formula:

y =

23 +/- SQRT[529 - 4(3)(30)]
----------------------------------
2(3)

=

23 +/- SQRT(529 - 360)
----------------------------
6

=

23 +/- SQRT(169)
----------------------
6

=

23 +/- 13
-------------
6

= 10/6, 6 = 5/3, 6

Now just take the one that produces the larger result for volume, since you want the maximum. Note that neither of these is larger than the dimensions of the original square, which would be unacceptable.

Actually, 6 is unacceptable because 2*6>8 (8 being the shorter dimension of the original rectangle).

Oops, right, so it's 5/3.

wow thanks all for helping me out!! I understand how to solve for x now hee thanks again!! If it's not too much bother, can offer advice on this question?

A pencil tub (open cylinder) has a surface area of 49/3 * pi

What is the radius of the tub to make the volume a max?

So what I have done is:

Generate a formula for the area = pi * r^2 + 2* pi * r * h = 49/3 * pi

volume = pi * r^2 * h

When I solved the area for one variable, I got h = 49/r - (1/2)*r....is that right? I'm not very good at manipulating equations.....sigh

I tried to quote myself to correct an arithmetic error and ended up deleting most of this post. I'll try to reproduce it and correct the calculation mistake at the same time:

pi * r^2 + 2* pi * r * h = 49/3 * pi
r^2 + 2rh = 49/3
2rh = 49/3 - r^2
Dividing through by 2r to isolate h:
h = 49/6r - r/2

Now insert this in the formula for volume and set the derivative equal to zero to find the maximum. Note that you could have left the r and h terms mixed and carried dh/dr through your equations, but your approach is a good one.

Aww man, I must've manipulated my equations wrong........I ended up with like no r terms when I differentiated and set the equation to 0 ..............

V = 49/6 * pi * r ^3 - pi * r ^3

dv/dr = 147/6 * pi * r ^ 2 - 3* pi * r ^2

and I set that to zero to find r, so by putting 3 * pi * r ^2 = 147/6 * pi * r ^2 ( should I take out a common factor of 3 btw from the fraction?)

Since there are pi * r ^2 terms on both sides, do they not cancel off? Ahh so confused

V = pi*(r^2)*h
h = 49/(6*r) - r/2
Therefore,
V = pi*(r^2)*[49/(6*r) - r/2]
= pi*[(r^2)*49/(6*r) - (r^2)*r/2]
= pi*[(49/6)*r - (1/2)*r^3]

Set the derivative of (49/6)*r - (1/2)*r^3 equal to 0.

*Updated to account for Brandon9000's fix

I made an arithmetic mistake, it's 49/6r - r/2.

o yer o yer dat makes sense now! Thanks Brandon9000 and markr

look at it this way
Surface area is
S=pi*r(r+2h)
dS=0=pi*[(r+2h)+r]dr+2pi*rdh
factor out 2pi & this becomes
(r+h)dr+rdh=0
rearranging this becomes
dh/dr=-(r+h)/r

And volume is given by
V=pi*r^2*h
again
dv=0=pi*r^2dh + 2pi*rhdr
Factoring ouit pi*r
rdh+2hdr=0
and dh/dr=-2h/r

Setting dh/dr=dh/dr
you get -(r+h)/r=-2h/r
cancelling -1/r you get
r+h=2h
so r=h
to maxinize

You know that S=pi*r(r+2h)=49/3*pi
and h=r
so pi*r(r+2r)=pi*3r^2=pi*49/3
and factoring pi/3
r^2=(49/3)/3
so
r^2=49/9
This is a perfect square 49/9=(7/3)(7/3)
so r=7/3 and since r=h, h=7/3

this should be the shape of the maximal volume pencil cup, as driven by convention.

Rap

oo thnx Rap!

First you have to get the relationship between the height and the base of the equilateral triangle. An equilateral triangle has three equal legs (say l). The height is the length of a segment that bisects one angle and the opposite leg. Moreover this makes two right triangles out of the equilateral triangle. Using pythageorus, the hypotenuse of this right triangle is l (the length of a side of the original triangle), the longer leg is h, and the shorter leg is l/2 (the bisected length of the original) triangle. So h^2+(l/2)^2=l^2. Now solve for the height of the triangle and h=sqrt(3)/2l.

Now find the area of the triangle from one half the base times the height. The base is l*, and the height is sqrt(3)/2l so the area of the equilateral triangle base is sqrt(3)/4*l^2.

* caution this is a variable-the letter ell(l) and not a constant--the number one(1)

Now this prism has a top and a bottom (two equalateral triangles) and three rectangular sides to make five sides. So the prism is closed.

The surface area of the sides of the prism is the product of length of the equilateral triangle leg (l) and the height (this is a variable I'll call x) so the surface are of a side is lx

The surface area of the prism is 2 times the area of the equilateral triangles and 3 times the area of a rectangular side.

This becomes
S= 2At +3Side or
S=2*sqrt(3)/4*l^2+3lx

The prism volume becomes the area of the base times the height (x) or
V=At*x=sqrt(3)/4*l^2*x

Now do the same as the pencil cup differentiate the surface area and volume relations with respect to l and x and set to zero (to get some maxima or minima). Solve each relationship in terms of dl/dx (or vice versa) and set equal to each other. this yields a relationship between l and x for an optimum closed prism.

Substitute this relationship into the prism volume formula, relate to the specified volume and solve for the (now single) unknown.

Clue---when I work through it all I get that the optimal height (x) is equal to the product of sqrt(3) and the length of the leg of the equilateral triangle (l). The cube of the height when (given the volume) is 16/9

Rap

ok I got up to finding equations for volume and area but I'm stuck at the differentiating part nowz. Raprap, the method you use is the one I wasn't taught to use, is there an alternative? Can I solve volume for one variable and then sub it into the surface area equation?

Sore you do

look at your entries on differentials and you should dee something that should look like

d(f(x)g(x))/dx=f(x)g'(x)+f'(x)g(x)

In a way it's analogous to say the volume relationship

of your pencil cup (cylinder)

where V=pi*r^2*h

so is you want the maximum or minimum of volume with changing r's and h's and let f(r)=r^2 and g(h)=h

V=pi*f(r)*g(h)

so differentiate V with respect to the independent variables (r and h) and set the derivative equal to zero

dV=pi[f'(r)g(h)+f(r)g'(h)]=0

and with substitution the derivities become

f'(r)=2rdr and g'(h)=dh

putting this back into a zero valued dV

0=pi[2hrdr+r^2dh]

and factor out pi*r

2hdr+rdh=0

As for substitution of h into r the other equation and vice versa, no that doesn't work--look as volume and surface areas as dependant variables. At some point they are have a maximum or a minimum (their derivative is zero). It is at this derivative your able to find a commonality between the other two independent variables. It is at that point you can find a relationship between these independent variables.

Rap

ah k I'll give it a try thnx raprap.

The other part of this question is that once I've found the answer to this question, how would I generate a formula using V instead of the given 4/square root of 3? Like so that if I plug in any values I can calculate using this so-called general formula? I've no idea how to do this question. Any suggestions?

Also here's a riddle type question:

It says a school is holding a charity event and tickets for pupils cost 2.00 whilst tickets for teachers cost 3.00. If a total of 250 people bought tickets and 580.00 of ticket money was collected, then how many of those were pupils? I managed to solve this using two variables x and y and I got the answer 170 pupils but how would I solve using just the one variable? Any suggestions?

You can't. You can eliminate one variable, but you've still got two===fortunately you've got two equations. Look at it this way, in a Euclidian space (a flat plane in two dimensions) you can express a line by y=mx+b. Your ticket problem is two lines?-they can cross (and you get a single solution, they're parallel and you get solutions, or they're the same line. So if I were to take your two ticket solutions with (X) being the students and (Y) the teachers. The total sold is 250, and if the students cost is $2 and the teachers $3 and $580 was collected.

Tickets sold 250 is the number of students (X) and (+) the number of Teachers (Y) or
250=X+Y
but hey let's look at this as an equation in the form of a line then
Y=mX+b
then m= -1 and b=250
so
Y= -X+250

Now the money collected is $580 or $2 times the number of students (X) and (+) $3 times the number of teachers (Y). This becomes
$580=$2X+$3Y
if I put this in the form of a line
Y=mX+b
Then m= -2/3
And b=$580/3 (I'll leave this as an improper fraction)
So
Y= -2/3X+580

I've still got two equations, but they're still both equal to the same thing (Y)
So
-X+250= -2/3X+580/3
and multiplying both sides through by 3 yields
-3X+750= -2X+580
a quick rearrangement gives X=(750-580)=170 and X is the number of students.

Then how many teachers were there, that's easy Y= -170+250=80 teachers.

What's neat is if you plot both of these lines on graph paper they'll cross at a point (x,y), and that point will be (170,80)

Ok the differential problems?-look you're doing the same thing, you're just using a property of what it means when the 1st differential is zero. Go back to the line (Y=mX+b) and look at the differential of Y with respect to X (dY=mdX). .It's the slope. As a consequence all lines (with the exception of horizontal and vertical lines) have a slope that is between 0 and infinity. Higher order equations can have places where the slope changes and some have places where the slope is zero?-

What you're doing with these two relationships is using that place where the slopes are the same. You're taking the slope of two relationships and finding that place where the slopes are zero (this is why you find the differential and set it to zero). You then investigate the relationship between two (note the two) independent variables at those points with a zero slope for both equations.

My recommendations on solving one of these problems (note a lot of this always applies)

Make sure I have as many dependent relations as I have unknowns (this always applies)

Create good models for dependant variable (Surface Area and Volume) in terms of independent variable (diameter or side length, height)

Differentiate both of the Dependant variable in terms of both independent variables [f'(xy)=xy'+x'y} Set both the Dependant differentials to zero (a max/min) find the relationships between x'/y' or y'/x' for both relations.

Rearrange to separate the independent variables. So now I know if I have a can and I want to use a minimal amount of material and hold the maximum amount of product, the relation between height and sides is this.

BTW this is a practical packaging problem---You create a model that calculates say the cost of a container based upon some dimensions (independent variables) and capacity of that container with potential constraints (boxes can get too big and heavy) and look for an optimum (usually least cost).

Rap