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MCV4U1-UNIT THREE Lesson Three

Lesson 3: OPTIMIZATION PROBLEMS

Recall: A max or min occurs when the first derivative is zero.

1. If 2700 square centimeters is available to make a box with a square base, find possible volume of the box. h

*SA 2 x 2 4 xh*

*2700 2 x 2 4 xh*

1350 x 2 2 xh and V x 2 h the largest

x x

We need the quantity that is being maximized (the volume) to be in terms of one variable. Since it would be easier to solve for h in the SA formula, we will write the volume in terms of x.

*1350 x 2 2 xh*

*1350 x 2 2 xh*

1350 x 2

*2 x*

* h*

V x 2

1350 x 2

*2 x*

V

1350 x 2

*2 x*

V 675 x

1

2 x 4

2 x x 3

* dV dx dV*

675

3

*2 x 2 dx*

3

2 x 2

0 when 675

675 x 2 450 x 21.2

3

2 x 2 0

But, as with our other extrema problems, we need to check the boundary points in the interval.

What is the interval here?

We know that.....

1

*MCV4U1-UNIT THREE Lesson Three x 0 and h 0, so*

*1350 x*

*2 x*

2

0

1350 x 2 0 x 2 1350 x 1350 ( 36.7) x 1350

0 x 1350

Now we must evaluate V(x) for the values, x 0, x 450, x 1350

*V*

0

*V*

9545.94

*V*

1350

0

Therefore, the largest possible volume of the box is 9545.94 cm 3

2. least amount of fence she will need river?

A farmer wishes to fence in a rectangular pasture on a 3750 square metre piece of riverfront regions, as illustrated below. What is the to purchase, assuming that she will not erect a fence along the x y

2