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labwork [276]
2 years ago
5

Phone Company A charges a monthly fee of $42.50, and $0.02 for each minute talk time. Phone company B charges a monthly fee of $

25.00, and $0.09 for each minute of talk time.
Mathematics
1 answer:
Y_Kistochka [10]2 years ago
8 0
We need to call for x minute
+ Phone Company A charges a monthly fee of $42.50, and $0.02 for each minute talk time. So we have to spend: <span>$42.50+ $0.02x
+ </span>Phone company B charges a monthly fee of $25.00, and $0.09 for each minute of talk time. So we have to spend: <span>$25.00+ $0.09x

We solve for x: </span>$42.50+ $0.02x> <span>$25.00+ $0.09x
or </span>$42.50- $25.00 > $0.09x- <span>$0.02x
and we have $0.07x<$27.50
or x< 27.50:0.07 and x< 393.86

The answer is:
If we have to call much time, at least 394 minutes, we should choose A
If not, choose B</span>
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Let's consider a simple example of estimating the value of a general definite integral,

\displaystyle\int_a^bf(x)\,\mathrm dx

Split up the interval [a,b] into n equal subintervals,

[x_0,x_1]\cup[x_1,x_2]\cup\cdots\cup[x_{n-2},x_{n-1}]\cup[x_{n-1},x_n]

where a=x_0 and b=x_n. Each subinterval has measure (width) \dfrac{a-b}n.

Now denote the left- and right-endpoint approximations by L and R, respectively. The left-endpoint approximation consists of rectangles whose heights are determined by the left-endpoints of each subinterval. These are \{x_0,x_1,\cdots,x_{n-1}\}. Meanwhile, the right-endpoint approximation involves rectangles with heights determined by the right endpoints, \{x_1,x_2,\cdots,x_n\}.

So, you have

L=\dfrac{b-a}n\left(f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1})\right)
R=\dfrac{b-a}n\left(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n)\right)

Now let T denote the trapezoidal approximation. The area of each trapezoidal subdivision is given by the product of each subinterval's width and the average of the heights given by the endpoints of each subinterval. That is,

T=\dfrac{b-a}n\left(\dfrac{f(x_0)+f(x_1)}2+\dfrac{f(x_1)+f(x_2)}2+\cdots+\dfrac{f(x_{n-2})+f(x_{n-1})}2+\dfrac{f(x_{n-1})+f(x_n)}2\right)

Factoring out \dfrac12 and regrouping the terms, you have

T=\dfrac{b-a}{2n}\left((f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1}))+(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n))\right)

which is equivalent to

T=\dfrac12\left(L+R)

and is the average of L and R.

So the trapezoidal approximation for your problem should be \dfrac{14+21}2=\dfrac{35}2=17.5\text{ in}^2
4 0
2 years ago
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