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Answer:
The dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches
Step-by-step explanation:
We have that:
Let the dimension of the paper be x and y;
Such that:
So:
Substitute 128 for Area
Make x the subject
When 1 inch margin is at top and bottom
The length becomes:
When 2 inch margin is at both sides
The width becomes:
The New Area (A) is then calculated as:
Substitute for x
Open Brackets
Collect Like Terms
To calculate the smallest possible value of y, we have to apply calculus.
Different A with respect to y
Set
This gives:
Collect Like Terms
Multiply through by
Divide through by 2
Take square roots of both sides
Recall that:
Recall that the new dimensions are:
So:
To double-check;
Differentiate A'
The above value is:
This means that the calculated values are at minimum.
<em>Hence, the dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches</em>
SOLUTION:
Step 1:
In this question, we are given the following:
Step 2:
In the question, we are looking at the absolute value function which says that:
Based on this, we have that:
Answers:
x = angle MLN = 35 degrees
angle FLJ = 55 degrees
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Explanation:
This problem is fairly tricky if you're not sure what to look for. A slight clue is that they've marked a red point that strangely doesn't have a label on it. It's a fairly small point but it's definitely there if you look closely. While this point's location isn't exactly what we want, we're fairly close. Start at point L and draw a ray through the center point P. Ray LP will intersect the circle at point A, as shown in the diagram below.
From here, draw segments FA and MA. Now notice that inscribed angle LMF = 55 and inscribed angle FAL both subtend the same arc. The term "subtend" basically means "cut off". This arc that the inscribed angles subtend is minor arc FL. A minor arc is where you travel the shorter path around the circle, which indicates its measure is less than 180 degrees.
Since inscribed angles LMF and FAL subtend the same minor arc, this makes the inscribed angles to be congruent.
In short: angle FAL is 55 degrees
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Segment LA goes through the center P. Through Thales theorem, we know that inscribed angle LFA is 90 degrees. Consequently, we can determine that inscribed angle FLA is 90-55 = 35 degrees.
Segment JL is tangent to the circle, meaning that angle ALJ is 90 degrees. So angle FLJ is 90-35 = 55 degrees.
It's not a coincidence that angle FLJ, angle FAL, and angle LMF are the same measure.
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We found that angle FAL was 55 degrees. Applying Thales theorem again shows that angle MAF is 90 degrees. Therefore, angle LAM is 90-55 = 35 degrees.
Focus now on triangle LMA. This is also a right triangle (Thales Theorem). The upper acute angle we found was angle LAM = 35, so the lower acute angle is ALM = 55.
Then we can find angle MLN = (angle ALN) - (angle ALM) = 90 - 55 = 35
In short, angle MLN = 35 degrees
Similar to the previous section, it is not a coincidence that angles LFM, LAM and MLN are the same measure.
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As an alternative, since we know angle FLJ = 55, and angle FLM is 90 degrees, this means...
(angle FLJ)+(angle FLM) + (angle MLN) = 180
55 + 90 + angle MLN = 180
145 + angle MLN = 180
angle MLN = 180 - 145
angle MLN = 35 degrees
