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3 years ago

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Mrs. Kosko left $19.47 on the table in a restaurant after receiving the bill for $16.50. If there was no tax, what percent did s

he tip? (A) 15% (B) 13% (C) 17% (D) 18%

2 answers:

attashe74 [19]3 years ago

7 0

Take the cost of the bill and add the tip percentage of the bill. That is:

$cost = bill + (bill \times\% \: tip)$

or

$cost = bill(1 + \% \: tip)$

solve:

$19.47 = 16.50(1 + \% \: tip)$

$\frac{19.47}{16.50} = 1 + \% \: tip$

$1.18 = 1 + \% \: tip$

$\% \: tip = 0.18 = 18\%$

The tip was 18% of the bill.

spayn [35]3 years ago

6 0

She tipped 18 percent . 16.50x0.18= 2.97. 16.50+2.97= 19.47

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Notice that both $t = 1$ and $t = 4$ are first derivatives of $A^{\prime}(t)$ over the interval $0 \le t \le 6$ .

However, among these two zeros, only $t = 1\!$ ensures that the second derivative $A^{\prime\prime}(t)$ is smaller than zero (that is: $A^{\prime\prime}(1) < 0$ .) If the second derivative $A^{\prime\prime}(t)\!$ is non-negative, that zero of $A^{\prime}(t)$ would either be an inflection point (if $A^{\prime\prime}(t) = 0$ ) or a local minimum (if $A^{\prime\prime}(t) > 0$ .)

Therefore $\! t = 1$ would be the only local maximum over the interval $0 \le t \le 6\!$ .

Calculate the value of $A(t)$ at this local maximum:

$A(1) = 15.5$ .

Compare these three possible maximum values of $A(t)$ over the interval $0 \le t \le 6$ . Apparently, $t = 6$ would maximize the value of $A(t)\!$ . That is: $A(6) = 28$ gives the maximum value of $\! A(t)$ over the interval $0 \le t \le 6\!$ .

However, note that the maximum over this interval exists because $t = 6\!$ is indeed part of the $0 \le t \le 6$ interval. For example, the same $A(t)$ would have no maximum over the interval $0 \le t < 6$ (which does not include $t = 6$ .)

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