# Homework Help: Calculus maxima and minima word question can't understand

1. Aug 24, 2004

### singleton

Calculus maxima and minima word question...can't understand :(

A rectangular field is going to be enclosed and divided into two separate rectangular areas (not equal either). Find the minimum fencing required if the total area of the field is 1200m^2

(See the picture attached right now)

Let x represent the width of the rectangular area's width in metres
Let y represent the length of the first rectangular area in metres
Let z represent the length of the second rectangular area in metres

I've identified that we want to minimize the total amount of fencing P
So the function is:
P = 3x + 2y + 2z

Given the total area = area of first rectangle + area of second rectangle
1200 = xy + xz

The only problem is that I have three variables and I think I'm supposed to somehow do this with two as we've only been taught how to use two so far. I could figure it out with two but I don't know what is wrong with my answer so far :( Have I wrongly interpreted the question?

thanks for any help!!

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2. Aug 24, 2004

### singleton

I can't believe I did something so stupid!!! :(

I shouldn't even have y and z it should just be y

P = 3x + 2y

1200 = xy
x = 1200/y

P = 3(1200/y) + 2y
P = 3600/y + 2y

first derivative is -3600/y^2 + 2
second derivative is 7200/y^3
etc etc (I think)

blah I can't believe I did something so stupid :(

3. Aug 24, 2004

### NateTG

On the other hand, you appear to have the situation well in hand.