# MCV4U1-UNIT THREE Lesson Three Lesson 3: OPTIMIZATION PROBLEMS

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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

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

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

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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

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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

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