# Tricky word problems

• Jan 3rd 2008, 09:00 AM
tommyjay117
Tricky word problems
I'm having the worst time with these.

A man has 340 yards of fencing for enclosing two seperate fields, one of which is to be a rectangle twice as long as it is wide and the other a square. The square field must contain a least 100 square yards and the rectangular one must contain at least 800 square yards.

a) If x is the width of the rectangular field, what are the maximum and minimum possible values of x?

b) What is the greatest number of square yards that can be enclosed in the two fields? Justify your answer.

I don't even know what to do with calculus on this one. Do I need to create an equation for the width/size with x and then derive and find the minimum/maximum?

Here's another one that is similar.

The number of bacteria at time t is given approximately by y = 1000(25 + te^(t/20) for 0 ≥ t ≥ 100.

a) Find the largest number and the smallest number of bacteria in the culture during the interval.

b) At what time during the interval is the rate of change in the number of bacteria a minimum.
• Jan 3rd 2008, 10:11 AM
TKHunny
Quote:

Originally Posted by tommyjay117
A man has 340 yards of fencing for enclosing two seperate fields, one of which is to be a rectangle twice as long as it is wide and the other a square. The square field must contain a least 100 square yards and the rectangular one must contain at least 800 square yards.

a) If x is the width of the rectangular field, what are the maximum and minimum possible values of x?

b) What is the greatest number of square yards that can be enclosed in the two fields? Justify your answer.

I don't even know what to do with calculus on this one. Do I need to create an equation for the width/size with x and then derive and find the minimum/maximum?

What have you tried?

You have:

Rectangular:

X = Width
L = Length
Perimeter: 2X + 2L < 340
Area: X*L >= 800

Square:

S = Side
Perimeter: 4S < 340
Area: S*S >= 100

It should also be clear that:

2X + 2L + 4S = 340

Now a little leg-work...

Perimeter: 2X + 2L < 340 ==> X + L < 170
Area: X*L >= 800 ==> L >= 800/X
Ah, so X + 800/X < 170 <== This isn't entirely obvious. Look closely.

Perimeter: 4S < 340 ==> S < 85
Area: S*S >= 100 ==> S >= 10
Also, S = (340 - 2X - 2L)/4 = (170 - X - L)/2

Are we getting anywhere?

What happens to X and L if we fix S = 10? S = 85?
What happens to L and S if we fix X + 800/X = 170?

Let's see where this leads you.