# 幫我翻譯微積分數學 10點

or, more simply, dividing by 4, it is x3 − 4x = 0. Luckily, we can see how to factor this: it is

x(x − 2)(x + 2)

So the critical points are −2, 0, +2. Since the interval does not include −2, we drop it from our list.

And we add to the list the endpoints −1, 3. So the list of numbers to consider as potential spots for

minima and maxima are −1, 0, 2, 3. Plugging these numbers into the function, we get (in that order)

−2, 5,−11, 14. Therefore, the maximum is 14, which occurs at x = 3, and the minimum is −11, which

occurs at x = 2.

Notice that in the previous example the maximum did not occur at a critical point, but by coincidence did

occur at an endpoint.

You have 200 feet of fencing with which you wish to enclose the largest possible rectangular garden. What

is the largest garden you can have?

Let x be the length of the garden, and y the width. Then the area is simply xy. Since the perimeter is

200, we know that 2x + 2y = 200, which we can solve to express y as a function of x: we find that y =

100 − x. Now we can rewrite the area as a function of x alone, which sets us up to execute our procedure:

area = xy = x(100 − x)

The derivative of this function with respect to x is 100 − 2x. Setting this equal to 0 gives the equation

100 − 2x = 0

to solve for critical points: we find just one, namely x = 50.

Now what about endpoints? What is the interval? In this example we must look at ‘physical’ considerations

to figure out what interval x is restricted to. Certainly a width must be a positive number, so x > 0

and y > 0. Since y = 100 − x, the inequality on y gives another inequality on x, namely that x < 100. So

x is in [0, 100].

When we plug the values 0, 50, 100 into the function x(100 − x), we get 0, 2500, 0, in that order. Thus, the

corresponding value of y is 100 − 50 = 50, and the maximal possible area is 50 · 50 = 2500.

Rating

or, more simply, dividing by 4, it is x3 − 4x = 0.

或，更簡單的說，除以4，得到x3 − 4x = 0

Luckily, we can see how to factor this: it is x(x − 2)(x + 2)

很幸運地，我們可以看出如何將其因式分解：得到x(x − 2)(x + 2)

So the critical points are −2, 0, +2.

所以臨界點為−2, 0, +2

Since the interval does not include −2, we drop it from our list.

因為區間不包含−2，所以我們將−2排除在清單外

And we add to the list the endpoints −1, 3.

再來我們要將端點−1, 3放入我們的清單中

So the list of numbers to consider as potential spots for minima and maxima are −1, 0, 2, 3.

所以我們需考慮有可能的最大以及最小點在此數字清單中有−1, 0, 2, 3

Plugging these numbers into the function, we get (in that order) −2, 5,−11, 14.

將這些數字代回原式，我們可以分別得到2, 5,−11, 14

Therefore, the maximum is 14, which occurs at x = 3, and the minimum is −11, which occurs at x = 2.

所以，最大值是14發生在x = 3時，以及最小值是−11當x = 2時

2008-09-29 23:55:57 補充：

Notice that in the previous example the maximum did not occur at a critical point, but by coincidence did occur at an endpoint.

特別說明一下前面的例子，其最大值並不是發生在臨界點，而是非常巧合的發生在端點(指的是區間的兩個點)

2008-09-29 23:56:16 補充：

You have 200 feet of fencing with which you wish to enclose the largest possible rectangular garden.

你有200英呎的柵欄，而你希望圍成一個可能最大的矩形花園

2008-09-29 23:56:39 補充：

What is the largest garden you can have?

你能得到最大的花園有多大?

Let x be the length of the garden, and y the width.

假設x代表長度，以及y代表寬度

2008-09-29 23:57:23 補充：

Then the area is simply xy.

接著這個面積可以簡單地表示成xy

Since the perimeter is 200, we know that 2x + 2y = 200,

因為花園的周長是200，所以我們知道2x + 2y = 200

2008-09-29 23:57:49 補充：

which we can solve to express y as a function of x: we find that y =

100 − x.

我們可以得到一個x的函數來表示y：我們得到y =100 − x

Now we can rewrite the area as a function of x alone, which sets us up to execute our procedure: area = xy = x(100 − x)

現在我們可以改寫成單獨x的函數來表示面積，這可以使我們執行我們的步驟：面積= xy = x(100 − x)

2008-09-29 23:58:12 補充：

The derivative of this function with respect to x is 100 − 2x.

將這個函數對x作微分得到100 − 2x

Setting this equal to 0 gives the equation 100 − 2x = 0 to solve for critical points: we find just one, namely x = 50.

令上式為零得到一個式子100 − 2x = 0，並解出臨界點：我們找到一個，即x = 50

2008-09-29 23:58:38 補充：

現在端點是什麼?

What is the interval?

區間是多少?

In this example we must look at ‘physical’ considerations

to figure out what interval x is restricted to.

在此例中，我們必須考慮物理的因素才能計算出x被限制在什麼區間

Certainly a width must be a positive number, so x > 0 and y > 0.

的確，長度寬度必須是個正數，所以x > 0以及y > 0

2008-09-29 23:59:09 補充：

Since y = 100 − x, the inequality on y gives another inequality on x, namely that x < 100.

因為y = 100 − x，由y的不等式得到另一個x的不等式，即x < 100

So x is in [0, 100].

所以x是在[0, 100]這個區間中

2008-09-29 23:59:25 補充：

When we plug the values 0, 50, 100 into the function x(100 − x), we get 0, 2500, 0, in that order.

當我們把數值0, 50, 100代入函數x(100 − x)，我們分別得到0, 2500, 0

Thus, the corresponding value of y is 100 − 50 = 50, and the maximal possible area is 50 · 50 = 2500.

因此，y的相對應數值為100 − 50 = 50，而且最大可能的面積是50 · 50 = 2500

Source(s): 傻妞的好朋友