Answer:
Step-by-step explanation:
we are given an algebraic function in x, such that
To find when second derivative >0
For this let us find first and second derivative by successive differentiation.
when ii derivative >0 we have
60x^3-30>0
f"(x)>0 in
Answer:
Step-by-step explanation:
There are a number of ways to solve mixture problems like this. Some are simple enough you can almost write down the answer. One I like uses a diagram to help you find the ratios of the components of the mix. You can do the same thing with a formula (without the diagram).
The basic idea is that you compute a ratio between two differences. The numerator is the difference between the average interest rate and the lower interest rate. Here, the average interest rate is 2700/30,000 = 0.09 = 9%. Then the numerator is 9% -8.5% = 0.5%
The denominator is the difference between the two interest rates of the investments: 10% -8.5% = 1.5%. And the ratio of these two values is ...
0.5%/1.5% = 1/3
This is the fraction of the total investment that was invested at the higher interest rate:
(1/3)×$30,000 = $10,000 . . . . was invested at 10%
Then the remaining amount, $20,000, was invested at 8.5%.
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If you would prefer an equation, you can let x represent the amount invested at the higher rate. (That is the most convenient assignment of variables, as it keeps all of the numbers positive.) Then (30,000 -x) is the amount invested at the lower rate. The total amount of interest is ...
10%(x) +8.5%(30,000 -x) = 2700
x(10% -8.5%) = 2700 -8.5%(30000) . . . . . subtract 8.5%(30,000)
x = (2700 -8.5%(30,000))/(10% -8.5%) . . . . divide by the coefficient of x
We have purposely left the expression unevaluated so that you can compare it to the math we described above. If you factor out 30,000 from the numerator of this fraction, you see the ratio that we described in words:
This tells us $10,000 was invested at 10%, and $20,000 was invested at 8.5%.
