A photograph, 8cm by 10cm, needs to be framed so that the area of the white space surrounding the photo is twice the area of the inlaid photo. There must be a minimum of 2cm between each side of the photo and the frame. Find the dimensions of the frame that minimize the perimeter of the frame:
a) 15.5cm by 15.5 cm
b) 12cm by 20cm
c) 17.1cm by 14cm
d) 12cm by 14cm
2007-05-17 05:39:20 · 3 answers · asked by Model Beauty 1 in Science & Mathematics ➔ Mathematics
Area of photograph = 8*10 = 80 sq. cm
Amount of white space = 2*80 = 160 sq.cm
Total area of frame = 160 + 80 = 240 cm.
Find the option in which the product of the numbers is 240 sq. cm.
The answer is b)12cm by 20cm
2007-05-17 05:44:16 · answer #1 · answered by Akilesh - Internet Undertaker 7 · 0⤊ 0⤋
say x is width and y is height... because the area of the white space must be twice that of the picture and the area of the picture is 80, we know that xy - 80 = 160, i.e. xy = 240
12x14 is not 240 so you can rule out d immediately
Now consider the other contraint: x must be greater than 12 cm and y must be greater than 14 cm (the dimensions of the photo plus 4 cm)
ABC all satisfy this condition, so just pick the one where the sum of x and y is the least, thus minimizing the perimeter
a) 31, b) 32, c)31.1
a) is the choice (notice b and c don't EXACTLY equal 240, but i assume this can be attributed to rounding error)
2007-05-17 05:52:24 · answer #2 · answered by emp211 3 · 0⤊ 0⤋
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2007-05-17 05:43:41 · answer #3 · answered by Anonymous · 0⤊ 0⤋