answer:a. To maximize profit with uniform pricing, BCY should set the price where marginal revenue (MR) equals marginal cost (MC). MR = MC 10,000 - 20Q = 5Q Solving for Q, we get: Q = 400 units Substituting Q back into the demand curve, we find the price: P = 10,000 - 10(400) P = 6,000 per unit Consumer surplus (CS) is the area below the demand curve and above the price line. In this case, it is a triangle with a base of 400 units and a height of 4,000 (10,000 - 6,000). CS = (1/2) * 400 * 4,000 CS = 800,000 b. With first-degree price discrimination, BCY can charge each customer their maximum willingness to pay. In this case, the price will equal the marginal cost for each unit: P = MC P = 5Q Solving for Q, we get: Q = 667 units Substituting Q back into the demand curve, we find the price: P = 10,000 - 10(667) P = 3,335 per unit c. With perfect price discrimination, BCY's total profit is equal to the sum of the prices charged to each customer, minus the total cost of production. Since we are ignoring fixed costs, the total profit is simply: Total Profit = (P1 + P2 + ... + Pn) - (MC1 + MC2 + ... + MCn) Total Profit = (5Q1 + 5Q2 + ... + 5Qn) - (5Q1 + 5Q2 + ... + 5Qn) Total Profit = 0 Therefore, with perfect price discrimination, BCY's total profit is zero. The additional consumer surplus captured by switching from uniform pricing to first-degree price discrimination is the difference between the consumer surplus under uniform pricing and the consumer surplus under first-degree price discrimination. Additional CS = CS (uniform pricing) - CS (first-degree price discrimination) Additional CS = 800,000 - 0 Additional CS = 800,000 By switching to first-degree price discrimination, BCY can capture an additional 800,000 in consumer surplus.

answer:a) Using the 19.2% depreciation rate for the third year of a five-year property under the MACRS, we calculate the depreciation for the business use of the truck: [ text{Depreciation for Business Use} = 19.2% times (6,000 times 0.7) = 806.4 ] As the truck was used for 171 days in 2012 before being sold, the prorated depreciation for the year would be: [ text{Prorated Depreciation} = 806.4 times frac{171}{365} approx 377.78 ] b) Sale value = 3,000 To find the net book value for realized gain/loss calculation: [ text{Net Book Value} = (6,000 times 0.7 times 0.192) + (6,000 times 0.3) - 840 - 1,344 - 377.78 + (6,000 times 0.3) = 3,438.24 ] Realized Loss: [ text{Realized Loss} = 3,438.24 - 3,000 = 438.24 ] For recognized gain/loss calculation (business portion only): [ text{Business Portion of Net Book Value} = 6,000 times 0.7 - 840 - 1,344 - 377.78 = 1,638.22 ] Recognized Gain: [ text{Recognized Gain} = 3,000 - 1,638.22 = 1,361.78 ] The recognized gain of 1,361.78 is of business character, as it relates to the commercial use of the truck.

answer:The linear approximation of {eq}f(x) = sqrt{x} {/eq} at {eq}x = 9 {/eq} can be calculated using the point-slope form: {eq}y - f(a) = f'(a)(x - a) {/eq} where {eq}a = 9 {/eq}. First, we find the derivative: {eq}f'(x) = frac{d}{dx}(sqrt{x}) = frac{1}{2}x^{-1/2} = frac{1}{2sqrt{x}} {/eq} So, at {eq}x = 9 {/eq}, we have: {eq}f'(9) = frac{1}{2sqrt{9}} = frac{1}{6} {/eq} Next, we evaluate the function at the point {eq}x = 9 {/eq}: {eq}f(9) = sqrt{9} = 3 {/eq} Now we can plug these values into the point-slope form: {eq}y - 3 = frac{1}{6}(x - 9) {/eq} To convert this to slope-intercept form {eq}y = mx + b {/eq}, we solve for {eq}y {/eq}: {eq}y = frac{1}{6}x - frac{1}{6} cdot 9 + 3 {/eq} {eq}y = frac{1}{6}x - frac{3}{2} + 3 {/eq} {eq}y = frac{1}{6}x + frac{3}{2} {/eq} Therefore, the linear approximation of {eq}sqrt{x} {/eq} at {eq}x = 9 {/eq} is: {eq}boxed{y = frac{1}{6}x + frac{3}{2}} {/eq}

answer:To prove that f_r(x) is continuous at an irrational number alpha, we must show that for any epsilon > 0, there exists a delta > 0 such that |x - alpha| < delta implies |f_r(x) - f_r(alpha)| < epsilon. Since alpha is irrational, f_r(alpha) = 0. We want to find delta such that for all x satisfying |x - alpha| < delta, |f_r(x)| < epsilon. If x is irrational, then f_r(x) = 0, and the condition is automatically satisfied. If x is rational and non-zero, say x = frac{p}{q} in lowest terms, then f_r(x) = frac{1}{q^r}. Since r ge 1, we have |f_r(x)| = frac{1}{q^r} le frac{1}{q}. Choose delta = frac{epsilon}{2}. Now, let x be a rational number with |x - alpha| < delta. Then, |x| le |x - alpha| + |alpha| < frac{epsilon}{2} + |alpha|. If |x| le frac{epsilon}{2}, we are done, as |f_r(x)| le frac{1}{q} le |x| < epsilon. If |x| > frac{epsilon}{2}, since delta = frac{epsilon}{2}, we have |x - alpha| < frac{epsilon}{2}, which implies that |x| < |alpha| + frac{epsilon}{2}. In this case, |x| < frac{epsilon}{2} + |alpha| < epsilon, which also ensures |f_r(x)| < epsilon. Thus, for any epsilon > 0, there exists a delta = frac{epsilon}{2} such that |x - alpha| < delta implies |f_r(x) - f_r(alpha)| < epsilon, demonstrating the continuity of f_r(x) at irrational numbers alpha.