Max.Min.

Try these...

1. Divide 12 into two parts so that their product is a maximum.

2. A piece of wire 24cm long is bent to form a rectangle. Find the dimensions of the rectangle which has the maximum area.

3. Two numbers differ by 6. If the square of the larger number exceeds twice the smaller number by an amount D, what is the maximum value of D?

Originally posted by elopawn
Try these...

1. Divide 12 into two parts so that their product is a maximum.

2. A piece of wire 24cm long is bent to form a rectangle. Find the dimensions of the rectangle which has the maximum area.

3. Two numbers differ by 6. If the square of the larger number exceeds twice the smaller number by an amount D, what is the maximum value of D?

1. 6 x 6 = 36
2. 6cm x 6cm
3. Infinty

maybe 3. would make more sense, if it was to determine the minimum for D, which is 11 (for x=1 and y=-5)

How about this:

split 120 into a sum of two numbers so that the product of the first number with the square of the second number will be max.

Originally posted by crazyblue
maybe 3. would make more sense, if it was to determine the minimum for D, which is 11 (for x=1 and y=-5)

How about this:

split 120 into a sum of two numbers so that the product of the first number with the square of the second number will be max.

40 and 80

with a max of 40*80^2 = 256000


Wow I actually used calc for something outside of school. Who would of thought?

Originally posted by crazyblue
maybe 3. would make more sense, if it was to determine the minimum for D, which is 11 (for x=1 and y=-5)

What about x=3, y=-3?

D=0

Originally posted by XanthosNZ
What about x=3, y=-3?

D=0

That yields D=15

Originally posted by richjohnson
That yields D=15

I missed the twice.

D tends to infinity