Q: Given the diagram of a rectangular field with brick walls on the top and bottom and fences on the sides, if the length is $$L$$ meters and the width is $$w$$ meters, which of the following expressions gives the total cost $$C of $$enclosing the field if brick costs $$\40$ per meter and fence costs $$\$$5$$ per meter?

$$C = 40 \times 2L + 5 \times 2w$$

Q: If the field must have a minimum width of $$5$$ meters, what is the minimum possible cost using the original cost function $$C(w) = 80 \dfrac{230}{w} + 10w$$ for $$w \geq 5$$?

$$C = 80 \dfrac{230}{5} + 10 \times 5$$

Question 1

Given the diagram of a rectangular field with brick walls on the top and bottom and fences on the sides, if the length is $$L$$ meters and the width is $$w$$ meters, which of the following expressions gives the total cost $$C of $$enclosing the field if brick costs $$\40$ per meter and fence costs $$\$$5$$ per meter?

Accepted Answer

$$C = 40 	imes 2L + 5 	imes 2w$$

Question 2

If the area of the field is fixed at $$230\ \mathrm{m}^2$$, and the width is $$w$$, what is the length $$L in $$terms of $$w$$?

Accepted Answer

$$L = \dfrac{230}{w}$$

Question 3

Using the area constraint $$Lw = 230$$, which of the following expresses the total cost $$C in $$terms of $$w$$ only?

Accepted Answer

$$C = 80 \dfrac{230}{w} + 10w$$

Question 4

What value of $$w$$ minimizes the total cost $$C(w) = 80 \dfrac{230}{w} + 10w$$? (You may leave your answer in terms of a square root.)

Accepted Answer

$$w = \sqrt{1840/10}$$

Question 5

Suppose the cost per meter for the fence increases to $$\10$ per meter, while the brick remains $$\$$40$$ per meter. Which of the following would be the new total cost function in terms of $$w$$?

Accepted Answer

$$C = 80 \dfrac{230}{w} + 20w$$

Question 6

If the width $$w is $$doubled, what happens to the length $$L$$ and the total cost $$C ($$using the original costs and area constraint)?

Accepted Answer

Length $$L is $$halved, and total cost $$C$$ increases.

Question 7

If the total cost for a field with width $$w = 10$$ meters is $$C = 80 \dfrac{230}{10} + 10 	imes 10$$, what is the total cost?

Accepted Answer

$$C = 1840 + 100 = 1940$$

Question 8

If the cost per meter for the brick wall is reduced to $$\20$ per meter, and the area remains 230$$\ \mathrm{m}^2$$, what is the new cost function in terms of w$?

Accepted Answer

$$C = 20 \dfrac{230}{w} + 10w$$

Question 9

If the field must have a minimum width of $$5$$ meters, what is the minimum possible cost using the original cost function $$C(w) = 80 \dfrac{230}{w} + 10w$$ for $$w \geq 5$$?

Accepted Answer

$$C = 80 \dfrac{230}{5} + 10 	imes 5$$

Question 10

If the total cost must not exceed $$\2000$, what is the largest possible width w$ (rounded down to the nearest integer) that can be used, using the original cost function C(w) = 80$$ \dfrac{230}{w} + 10w$?

Accepted Answer

$$w = 20$$

Optimization and Cost Analysis for a Rectangular Field

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