
(a)
To Find: The values of S(3),S(15),S(63)
(a)

Answer to Problem 57PPS
S(3)=30 , S(15)=50 , S(63)=70
Explanation of Solution
Given: S(a)=10+20log4(a+1)
Given: S(a)=10+20log4(a+1)
When a=3
S(a)=10+20log4(a+1)S(3)=10+20log4(3+1)S(3)=10+20log44S(3)=10+20S(3)=30
When a=15
S(a)=10+20log4(a+1)S(15)=10+20log4(15+1)S(15)=10+20log416S(15)=10+40S(15)=50
When a=63
S(a)=10+20log4(a+1)S(63)=10+20log4(63+1)S(63)=10+20log464S(63)=10+20log443S(63)=10+60S(63)=70
Hence, S(3)=30 , S(15)=50 , S(63)=70
(b)
To Interpret: The meaning of the function.
(b)

Answer to Problem 57PPS
$3 is spent on advertising, $30000 returned in sales.
$15 is spent on advertising, $50000 returned in sales.
$63 is spent on advertising, $70000 returned in sales.
Explanation of Solution
Given: S(a)=10+20log4(a+1)
Given: S(a)=10+20log4(a+1)
S(3)=30 : $3 is spent on advertising, $30000 returned in sales.
S(15)=50 : $15 is spent on advertising, $50000 returned in sales.
S(63)=70 : $63 is spent on advertising, $70000 returned in sales
(c)
To Draw: The graph of the function.
(c)

Answer to Problem 57PPS
Explanation of Solution
Given: S(a)=10+20log4(a+1)
Given: S(a)=10+20log4(a+1)
The graph of the function is
(d)
To Explain: Why the money spent in advertisement is less efficient as it is used in large amounts.
(d)

Answer to Problem 57PPS
Fixed increase in dollar advertisement increases sales by diminishing amount as advertisement increases.
Explanation of Solution
Given: S(a)=10+20log4(a+1)
Given: S(a)=10+20log4(a+1)
The graph of the function is
From the graph it is seen that as the fixed increase in dollar advertisement increases sales by diminishing amount as advertisement increases.
Chapter 8 Solutions
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