F.4 maths (functions)

2009-11-06 2:41 am

1) Three sides of a rectangular yard are surrounded by a fence, and a wall stands at the remaining side of the yard. The width of the yard is x m and the area enclosed is A m^2 where A=x(160-2x).
(a) Find the maximum area of the yard.
(b) When the area of the yard attains its maximum, does its width equal half its length? Explain briefly.

2) It is given that the sum of two integers is 12, and one of them is x.
(a) Express the product of these two integers in terms of x.
(b) Find the two integers such that their product is at the maximum.

回答 (1)

老爺子

2009-11-06 3:20 am

✔ 最佳答案

1)
A = x(160 - 2x)
A = 160x - 2x^ 2
A = -2(x^2 - 80x)
A = 2(40^2) - 2(x^2 - 80x + 40^2)
A = 3200 - 2(x - 40)^2

For any real value of x, 2(x - 40)^2 ≥ 0
And thus A = 3200 - 2(x - 40)^2 ≤ 0
Thus, the maxium value of A = 3200
Maximum area = 3200 m^2

(b)
When the area of the yard attains its maximum, 2(x - 40)^2 = 0
x = 40
The width = x m = 40 m
The length = 3200/x m = 3200/ 40 m = 80 m
Hence, the width equals half its length.

2.
(a)
One of the integers = x
Another integer = 12 - x
Product of these two integers = x(12 - x)

(b)
Product of the two integers
= x(12 - x)
= -(x^2 - 12x)
= 6^2 - (x^2 - 12x + 6^2)
= 36 - (x - 6)^2

For any real value of x, (x - 6) ≥ 0
and thus 36 - (x - 6)^2 ≤ 36
Hence, when the product is at the maximum, (x - 6)^2 = 0
x = 6
12 - x = 6

The two required integers are 6 and 6.
(The question does NOT require two different integers.)

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