# Homework Help: Calculus - Derivatives and Applications

1. Apr 17, 2004

### toosm:)ey

I have been given an assignment by my teacher, only I can't answer 3 of the questions.

1) A rectangle is expanding so that its length is always three times its width. The perimetre of the rectangle is increasing at a rate of 12cm/minute. Find the rate of increase in area of the rectangle when the perimetre is 80cm.

Here's what I have:

3x=y 2x+2y=80, Area=xy

2x+2(3x)=80
x=10, y=30, dP/dt = 12 dA/dt=?

If I take the derivative of a=xy, I need to know the derivative of x and y. But I am not given these. I do not know how to find these.

2) A rectangular area of 3200m^2 is to be fenced off. Two opposite sides will use fencing costing $2/metre, and the remaining sides will use fencing that costs$8/metre. Find the dimensions of the rectangle with the least costs.

3200=xy

y=3200/x

P=2x+2y
P=2x+2(3200/x)
P=2x+6400/x
P=2x+6400(x^-1)

Then I am stuck =(

3) A wooden chest is rectangular in shape with its length along the fron twice as its width. The top, front, and two sides of the chest are made of oak; the back and bottom are made from pine. Oak costs three times as much as pine. Find the dimensions that minimize of the cost of the chest if the volume is 0.25m^2.

I have no idea where to start. I know that V=xyh and x=2y, so V=2(y^2)h

Thanks for all your help! =D If this is in the wrong section, please move it.

2. Apr 17, 2004

1) P = 8x
dP/dt = 8dx/dt

No?

2) Why are you stuck? You just have a function you need to minimize now. Differentiate and set to zero like usual.

3) Treat this problem like #2. Turn those word expressions for price into mathematical ones.

3. Apr 18, 2004

### toosm:)ey

Hello and thanks again cookie! The Master of Calculus =D

I've completed questions 1 and 2, could you check those for errors please? Number 3 I am still stuck on.

1)

If P=8x, then dP/dt=8dx/dt. As dp/dt=12, then dx/dt = 3/2 ?

and if 2x+2y=p, then 2dx/dt + 2dy/dt = dp/dt

Meaning, 2(3/2) + 2dy/dt = 12, and dy/dt=9/2

A=xy

da/dt=(dx/dt)y+(dy/dt)x
da/dt=(1.5 x 30) + (4.5 x 10)
da/dt=90m^2/min

2) P=2x+6400(x^-1)

dp/dt=2 - 6400(x^-2)
0=2 - 6400(x^-2)
2=6400/(x^2)
x^2 = 3200
x= 56.57
y=56.57

Cost = x(price) + y(price)
c(x) = 56.57(2) + 56.57(8)
c(x) = 113.14 + 452.56
c(x) = $565.70 3) V=2(y^2)h I am hoping that I can get h in terms of y, somehow. However, from looking at the question, that seems near impossible. We are told that oak costs three times as much as pine, but I do not know how much oak I am using compared to pine. So, I need to find the area of oak, compared with the area of pine? Thanks again cookie! ~Smiley 4. Apr 18, 2004 ### cookiemonster #1 is right, although you took a slightly circuitous route to get to it. You could have simplified everything with A = 3x^2 and P = 8x So P = 80 = 8x, therefore x = 10 dP/dt = 8dx/dt, therefore dx/dt = 3/2 So dA/dt = 6x*dx/dt = 6*10*3/2 = 90 #2) I think you made a math error. I'm getting 226.27 as the price. But maybe I made a math error. I did it pretty quickly. #3) Remember that V = lwh = 2w^2h = .25. That there is the relation between width and height. Now you can make price as a function of only width, or only height, whichever you choose. To simplify the cost, let the price of pine = p. So the price of oak = 3p. So we get an expression for price, P = Sides*Oak + Front*Oak + Top*Oak + Back*Pine + Bottom*Pine P = 2Lh*3p + wh*3p + wL*3p + wh*p + Lw*p Now use your two formulas relating L to w and h to w to eliminate all the variables. You should get an equation in terms of one variable w, L, or H as well as p in there. But p is an unknown constant, so your answer must be in terms of p. Just minimize P the normal way, treating p as any other constant. Edit: I should note that P is the price function and p is the cost of pine. They are different, so it's case-sensitive. cookiemonster Last edited: Apr 18, 2004 5. Apr 18, 2004 ### toosm:)ey Thanks again =) 1) Good thinking... Saves a lot of time doing it your way. I never liked the chain rule anyway =) 2) My answer is not changing. Are you getting the same values for x and y, that is, the rectangle is a square? And are you doing the same calculations? Also when calculating the cost, there are 2 x sides, and 2 y sides, should they not be in the equation? If so, the cost function would be c(x) = 4x + 16y =$1131.37

3) Where did you get the relationship V = lwh = 2w^2h = .25 from? I understand that V = lwh = 0.25, but not 2w^2h.

If V=2(l^2)h, and V = 0.25, then h = 0.25/(2l^2)

Price = 6hwp + 3hlp + 3lw + hlp + lwp
Price = 6hwp + 4hlp + 3lw + lwp

2l = w, so

Price = 6h(2l)p = 4hlp + 3l(2l) + l(2l)p
Price = 16hlp + (6l^2) + (2l^2)p

Knowing that h = 0.25/(2l^2)

Price = 16lp*(0.25/(2l^2)) + (6l^2) + (2l^2)p
Price = 16p*l**0.25/(2l^2)) + (6l^2) + (2l^2)p
Price = (2p*l)/(l^2) + 6(l^2) + 2p(l^2)
Price = 2p/l + 6l^2 + 2p(l^2)
Price = 2p(l^-1) + 6l^2 + 2p(l^2)
P'(l) = -2pl + 12l + 4pl
P'(l) = 12l + 2pl
0 = 12l + 2pl
0 = 6l + pl
0 = l*(6+p)
l=0

Gah....! How can l = 0? There would be no box ='(

Last edited: Apr 19, 2004
6. Apr 19, 2004

In the problem it states that l = 2w.

Well, no box would certainly minimize the cost! There should be a second solution. One that's non-trivial this time.

I don't have time to check the math right now, but I'll do it when I get home (many hours from now...).