The top and bottom margins of a poster are 4 cm and the side margins are each 8 cm. if the area of printed material on the poste
r is fixed at 388 square centimeters, find the dimensions of the poster with the smallest area.
1 answer:
Illustrate the given problem. Refer to the diagram attached to aid our solution.
Area of bigger rectangle:
A = (L+8)(W+16)
Area of smaller rectangle (printed area)
388 = LW
From the second equation, we can express L in terms of W.
L = 388/W
Replace this to the first equation:
A = (388/W+8)(W+16)
A = 388 + 6208/W + 8W + 128
A = 6208/W + 8W + 516
Derive A with respect to W and equate to zero (calculus):
dA/dW = -6208/W² + 8 = 0
-6208/W² = -8
W² = -6208/-8 = 776
W = √776 = 27.86 cm
L = 388/27.86 = 13.93 cm
Thus, the smallest area would be:
A = (13.93 cm)(27.86 cm)
<em>A = 388.09 cm²</em>
Area of bigger rectangle:
A = (L+8)(W+16)
Area of smaller rectangle (printed area)
388 = LW
From the second equation, we can express L in terms of W.
L = 388/W
Replace this to the first equation:
A = (388/W+8)(W+16)
A = 388 + 6208/W + 8W + 128
A = 6208/W + 8W + 516
Derive A with respect to W and equate to zero (calculus):
dA/dW = -6208/W² + 8 = 0
-6208/W² = -8
W² = -6208/-8 = 776
W = √776 = 27.86 cm
L = 388/27.86 = 13.93 cm
Thus, the smallest area would be:
A = (13.93 cm)(27.86 cm)
<em>A = 388.09 cm²</em>
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Answer with Step-by-step explanation:
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Where t corresponds to the year 2000=0
a.
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b.In 2004
t=4
c.In 2013
t=13
Solution:
The function is given below as
To figure out
To do this , we will substitute x= 2x-1 in g(x)
Hence,
The composte function will be
Step 2:
To figure out the domain,
In mathematics, the domain of a function is the set of inputs accepted by the function.
Hence,
The domain of the function is