Answer:
4,250,000 riyals
Step-by-step explanation:
The total cost of drilling a 500-meter well is the sum of fixed costs (1,000,000 riyals) and variable costs.
By integrating the marginal cost function we get the variable costs function. Evaluating the variable costs function from 0 to 500 meters, gives us the total variable cost.
Total cost of drilling is:
Answer:
(a) -3/4
(b) -0.75
(c) -0.75
Step-by-step explanation:
It's a bit hard to tell what constitutes an "iteration" when using the bisection method to approximate a polynomial root. For the purpose here, we'll say one iteration consists of ...
Thus, the result of the iteration consists of a bracketing interval and the choice of one of the interval's ends as the solution approximation.
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(a) We observe that the graphs intersect in the interval (-1, 0). For the first iteration, we evaluate f(x)-g(x) at x=-1/2. This tells us the solution is in the interval (-1, -1/2). The x-value closest to the root is x=-1/2.
For the second iteration, we evaluate the function f(x)-g(x) at x=-3/4. This tells us the solution is in the interval (-1, -3/4). The x-value closest to the root is x=-3/4.
For the third iteration, we evaluate the function f(x)-g(x) at x=-7/8. This tells us the solution is in the interval (-7/8, -3/4). The x-value closest to the root is x=-3/4.
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(b) The graph tells us the solution is approximately 0.7549. Rounded to 2 decimal places, the solution is approximately 0.75.
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(c) The above solution found after 3 iterations rounded to 2 decimal places is exactly 0.75.
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See the attached table for function values.
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<em>Comment on bisection iteration</em>
Since you cut the interval containing the root in half with each iteration, you gain approximately one decimal place for each 3 iterations. When the function value is very nearly zero at one of the interval endpoints, it can take many more iterations to achieve a better result.
Here, it takes 4 more iterations before an x-value becomes closer to the solution (x≈-97/128). And it takes one more iteration to move the end of the interval away from -3/4. After these 5 more iterations (8 total), the solution is known to lie in the interval (-97/128, -193/256). The corresponding solution approximation is -193/256. It is still only correct to 2 decimal places.
