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vazorg [7]
3 years ago
13

Jodi is cutting out pieces of paper that measures 8 1/2 inches by 11 inches from a larger sheet of paper that has an area of 100

0 square inches. What is the greatest possible number of pieces of paper that Jodi can cut out of the larger sheet?
Mathematics
1 answer:
vodomira [7]3 years ago
8 0
The sizes of the pieces of paper are all the same and fixed.  Most likely you'll get the max number of pieces by cutting out pieces with their edges parallel:
top of one parallel to and immediately below (no offset to the left or right) the previous piece. 

We know that the total area of the "mother" piece of paper is 1000 sq in and that the formula for area here would be A = x y, where x is the width of the "mother" piece and y is the height of the "mother piece."

Let's first look at the height, y, of the "mother" piece.  Assume that we're cutting 8.5 by 11 pieces with their 11" sides extending vertically, and their 8.5" sides horiz.

We want to maximize the area taken up by the pieces cut from the "mother" piece.  This area is    x * y = x * (x/8.5) times (y/11).  We use the fact that xy=1000 and solve it for y:  y = 1000/x.

Then x*(x/8.5)(1000/x) is to be maximized.

 x^2    1000
----- * --------- is to be maximized.  Unfortunately:  That comes to 1000x/8.5, which is not very helpful.


Let's try a different approach.  Let y be the height of the "mother" piece and x the width.  Then, as before, xy = 1000.

Let n be the number of pieces of paper 8.5 by 11.  Then y=11n and x=8.5n.

Let's let x=1000/y.  Then    y=11n and x = 1000/y, or 1000/[11n]).  We
want to maximize n, the number of pieces of paper.  Let's see if this "works:"

Maximize A = xy = (11n) * 1000/


Sorry.  I'm not able to move further forward at the moment.  Hope this discussion is of some use to you.

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3 years ago
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Answer:

Area of the remaining triangle with the villager is 1243.13 m²

Step-by-step explanation:

Triangle ABC is the triangular plot of a villager shown in the figure attached.

Sarpanch requested the villager to donate land which is 6 m wide and along the side AC which measures 132.8m.

Other sides of the plot has been given as AB = 50m and BC = 123 m.

Now area of this land before donation = \frac{1}{2}\times {\text{Height}}\times \taxt{Base}

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After donation of the land the triangle formed is ΔDBE.

In ΔABC,

tan(ABC)=(\frac{AB}{BC})

tan(∠ABC) = \frac{50}{123}

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∠ABC = tan^{-1}(0.4065)

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In ΔEFC,

tanC = \frac{EF}{CF}

0.4065 = \frac{6}{CF}

CF = \frac{6}{0.4065}

CF = 14.76 m

Since DE = AC - (CF + AG)

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               = 132.8 - 29.52

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Now in ΔDBE,

sin(∠DEB) = \frac{BE}{DE}

sin(22.12) = \frac{BE}{102.48}

DB = 102.48×0.3765

     = 38.59 m

Similarly, cos(22.12) = \frac{BE}{DE}

0.9264 = \frac{BE}{102.48}

BE = 102.48×0.9264

     = 94.94m

Now area of ΔDBE = \frac{1}{2}(DB)(BE)

                                = \frac{1}{2}(38.59)(94.94)

                                = 1831.87 square meter

Area of remaining triangle with the villager = Area of ΔABC - Area of ΔDBE

= 3075 - 1831.87

= 1243.13 square meter

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