
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