Lesson 18 Applied Maximum and Minimum Problems

LESSON 18 APPLIED MAXIMUM AND MINIMUM PROBLEMS

Examples Solve the following problems. A diagram may be used to identify your variable(s).

1. A closed box is to be constructed having a volume of 96 ft. The length of the bottom (and hence, the top) of the box is four times the width of the bottom (and hence, the top). Find the dimensions that require the least amount of material.

NOTE: The amount of material, which is needed to construct the top and bottom of the box, is given by square feet.

NOTE: The amount of material, which is needed to construct the front and the back of the box, is given by square feet.

NOTE: The amount of material, which is needed to construct the left-side and the right-side of the box, is given by square feet.

Thus, the amount of material, M, which is needed to construct this closed box, is given by square feet. Simplifying, we have that in square feet.

NOTE: M is a function of two variables x and y. In order to get M as a function of one variable in x or y, we will need to get a relationship between x and y. A relationship between x and y is an equation containing only the variables of x and y. We haven’t used the information that the volume of the box is to be 96 cubic feet. Since the volume of a box is given by the formula , then the volume of our box is . Thus, in order for the volume of our box to be 96 cubic feet, we need that . This is our relationship between x and y.

Now, we can solve the equation for x or y. It is easier to solve for y in terms of x. Thus, .

Since and , then =

= in square feet.

Thus, , where . We want to minimize M.

Since , then =

Critical Number:

We will use the Second Derivative Test to see if a local minimum occurs when .

Since = , then

= .

Thus, a local

minimum is occurring at . Thus, M is a minimum when

feet.

Since , then .

Since and , then = =

Answer: ft by ft by ft

2. An open rectangular box is to be constructed having a volume of 288 in. The length of the bottom of the box is three times the width of the bottom. The material for the bottom of the box costs 8 cents per square inch and the material for the four sides costs 5 cents per square inch. Find the cost of the box which is the cheapest to construct.

NOTE: An open box means no top.

NOTE: The amount of material, which is needed to construct the bottom of the box, is given by square inches. Thus, the cost to construct the bottom of the box is given by cents.

NOTE: The amount of material, which is needed to construct the front and the back of the box, is given by square inches. Thus, the cost to construct the front and the back of the box is given by cents.

NOTE: The amount of material, which is needed to construct the left-side and the right-side of the box, is given by square inches. Thus, the cost to construct the front and the back of the box is given by cents.

Thus, the cost, C, to construct this open box, is given by

in cents. Simplifying, we have that = = in cents.

NOTE: C is a function of two variables x and y. In order to get C as a function of one variable in x or y, we will need to get a relationship between x and y. A relationship between x and y is an equation containing only the variables of x and y. We haven’t used the information that the volume of the box is to be 288 cubic inches. Since the volume of a box is given by the formula , then the volume of our box is . Thus, in order for the volume of our box to be 288 cubic inches, we need that . This is our relationship between x and y.

Now, we can solve the equation for x or y. It is easier to solve for y in terms of x. Thus, .

Since and , then =

= = in cents.

Thus, , where . We want to minimize C.

Since , then =

Critical Number:

We will use the Second Derivative Test to see if a local minimum occurs when .

Since = , then

= .

Thus, a local

minimum is occurring at . Thus, C is a minimum when inches.

Now, we need to find the cost of constructing the box when inches. Since , then =

= =

Answer: (in cents)

3. Bill can only afford to buy 100 yards of fencing. If he uses the fencing to enclose his rectangular garden, then find the largest area that he can enclose.

Area of the enclosure:

Since the amount of the fencing is 100 yards, we have that . Solving for y, we have that .

Since and , then = .

Thus, where , . We want to maximize A.

Critical Number: 25

We will use the Second Derivative Test to see if a local maximum occurs when .

Since , then

Thus, a local maximum is occurring at . Thus, A is a maximum when yards.

Now, we need to find the area of enclosure when yards. Since , then

Answer: 625 yd

4. A company wants to use fencing to enclose a rectangular region of 750 square yards next to a warehouse. If they do not fence in the side next to the warehouse, what are the dimensions which would require the least amount of fencing?

Warehouse

Amount of fencing:

Since the area of the enclosure is 750 square yards, we have that . Solving for x, we have that .

Since and , then , where . We want to minimize F.

Critical Numbers: since because it is a dimension

We will use the Second Derivative Test to see if a local minimum occurs when .

Since , then

Thus, a local minimum is occurring at . Thus, F is a minimum when yards.

Since and , then = = =

Answer: yds by yds

NOTE: If we were given the function above without the information that y is a dimension, then the domain of F is the set . Then the critical numbers of F would have been . Since , then a local maximum occurs at .

5. A farmer wishes to fence a rectangular pasture with a total area of 1000 yd, and he wants to divide it into two parts with a fence across the middle. Fencing around the outside costs $7.00 per yard, but he can use less expensive fencing at $4.00 per yard as a divider. Find the dimensions of the region which will cost the minimum to construct.

COMMENT: Some students wonder how to run the divider. You can run the divider parallel to the width of the rectangular or parallel to the length of the rectangular. Your answer will tell you which way to run the divider for the construction.

NOTE: The cost of constructing the outside of the enclosure is given by .

NOTE: The cost of constructing the divider for the enclosure is given by .

Thus, the cost, C, to construct this enclosure, is given by

in dollars. Simplifying, we have that = in dollars.

NOTE: C is a function of two variables x and y. In order to get C as a function of one variable in x or y, we will need to get a relationship between x and y. A relationship between x and y is an equation containing only the variables of x and y. We haven’t used the information that the area of the enclosure is to be 1000 square yards. Since the area under the divider is essentially zero, then the area of the enclosure is . Thus, in order for the area of the enclosure to be 1000 square yards, we need that . This is our relationship between x and y.

Now, we can solve the equation for x or y. Solving for x in terms of y, we have that .

Since and , then =

= in dollars.

Thus, , where . We want to minimize C.

Since , then =

Critical Numbers: since because it is a dimension

We will use the Second Derivative Test to see if a local minimum occurs when .

Since , then

Thus, a local minimum is occurring at . Thus, C is a minimum when yards.

Since and , then = =

Answer: yds by yds

6. A rectangle is inscribed in an isosceles triangle whose base is 48 feet and sides are 30 feet. If two of the vertices of the rectangle lie on the base of the triangle, then find the dimensions of the largest rectangle that can be inscribed in the triangle.

NOTE: We will take advantage of the symmetry in the isosceles triangle and let 2x be the length of the rectangle that is inscribed in the triangle.

30 30 30

y y

2x x

48 24

NOTE: The number 18, which is circled was found by using the Pythagorean Theorem.

The area of the inscribed rectangle is given by . We need to get a relationship between x and y. This relationship comes from the similar triangles that we have above. Thus, we have . Solving for y, we have that .

Since and , then =

= in square feet.

Thus, , where . We want to maximize A.

= =

Critical Number: 12

We will use the Second Derivative Test to see if a local maximum occurs when .

Since , then .

Thus, a local maximum is occurring at . Thus, A is a maximum when feet.

Since the length of the rectangle is , then the length is 24 feet. Since the width of the rectangle is , then the width is feet.

Answer: 24 ft by 9 ft

7. Find the volume of the largest cylinder that can be inscribed in a cone with height 12 meters and base radius 5 meters.

Since the volume of a cylinder is given by . We want to maximize V. We need to obtain a relationship between r and h. In order to obtain this relationship we will look at a cross section of the three-dimensional picture. Note that the cross section of the cone is an isosceles triangle and the cross section of the cylinder is a rectangle. Thus, the cross section looks like the following:

Using similar triangles, we have that . Solving for h, we have that .

Since and , then =

= in cubic meters.

Thus, , where . We want to maximize V.

Critical Number: since because it is a dimension

We will use the Second Derivative Test to see if a local maximum occurs

when .

Since , then = .

Thus, a local maximum is occurring at . Thus, V is a maximum when meters.

Since , then =

Answer: m