Question

求救，这道题怎麼做？

Answer 1

英文太差，麻烦楼主自己翻译吧。解答如下：
1）矩形宽为X-2，长为X-2+6=X+4，故周长=2×[（X-2）+（X+4）]=4X+4（周长≤33，所以4X+4≤33，X≤7.25）；
2）可知矩形宽度必大于0，即X-2>0，所以X>2，故命题为真；
3）可知X∈（2，7.25]，矩形面积=长×宽=（X-2）（X+4）=X²+2X-8，设为y，y>0，可知该二次函数开口向上，对称轴为X=-1，当X=-4和2时，y为0。所以y在区间（2，7.25]上单调递增且为正。矩形最大面积为y|X=7.25=5.25×11.25=59.0625。

Answer 2

(1)
Since the length of the rectangle is 6 cm longer than the width and the width is (x-2) cm, the length of the rectangle is (x-2)+6=(x+4) cm.
Hence, the perimeter of the rectangle is 2*(x-2+x+4)=(4x+4) cm
(2)
Yes, I agree. If x is less than 2, the width of the rectangle is x-2 which is less than 0. However, the width of a rectangle should be a positive number. So I agree with Annie.
(3)
Since the perimeter of the rectangle is not greater than 33 cm, we have 4x+4≤33.
By solving the inequality, we get x≤29/4=7.25. Due to x is an integer and x>2, so x could be 3,4,5,6,7.
Since the length of the rectangle is (x+4) cm and the width is (x-2) cm, the length and the width both become longer as x becomes larger, so does the area of the rectangle.
Hence, it is easy to see that when x=7, the area of the rectangle is maximized. So the maximum area of the rectangle is (7+4)×(7-2)=11×5=55 cm².

Answer 3

a. Let W = x-2, then L = W+6 = x+4 So, we have W+L = 2x+2
and P = 2(W+L) = 4x+4 < or = 33
b. Yes, I agree because the width must be larger than zero.
c. The largest integer x will give the maximum area. Therefore, x = 7, and the maximum area = (x-2)(x+4) = (7-2)(7+4) = 55 cm^2