answer:1. Given the equation: [ frac{a}{1-x} + frac{b}{1-y} = 1 ] where (a, b) are positive real numbers and (x, y) are positive real numbers less than 1. 2. We need to prove that: [ sqrt[3]{ay} + sqrt[3]{bx} leq 1 ] 3. We will use the Arithmetic Mean-Geometric Mean (AM-GM) inequality, which states that for non-negative real numbers (u_1, u_2, ldots, u_n): [ frac{u_1 + u_2 + cdots + u_n}{n} geq sqrt[n]{u_1 u_2 cdots u_n} ] with equality if and only if (u_1 = u_2 = cdots = u_n). 4. Consider the terms in the given equation: [ frac{a}{1-x} + frac{b}{1-y} = 1 ] We can rewrite this as: [ frac{a}{1-x} + frac{b}{1-y} + (1-x) + (1-y) = 2 ] 5. Applying the AM-GM inequality to the four terms (frac{a}{1-x}, frac{b}{1-y}, 1-x, 1-y): [ frac{frac{a}{1-x} + frac{b}{1-y} + (1-x) + (1-y)}{4} geq sqrt[4]{frac{a}{1-x} cdot frac{b}{1-y} cdot (1-x) cdot (1-y)} ] 6. Substituting the given equation into the inequality: [ frac{1 + (1-x) + (1-y)}{4} geq sqrt[4]{frac{a cdot b cdot (1-x) cdot (1-y)}{(1-x) cdot (1-y)}} ] Simplifying, we get: [ frac{2}{4} geq sqrt[4]{ab} ] [ frac{1}{2} geq sqrt[4]{ab} ] [ 1 geq sqrt[3]{ab} ] 7. Now, we need to show that: [ sqrt[3]{ay} + sqrt[3]{bx} leq 1 ] 8. Using the AM-GM inequality again for the terms (sqrt[3]{ay}) and (sqrt[3]{bx}): [ frac{sqrt[3]{ay} + sqrt[3]{bx}}{2} geq sqrt[3]{sqrt[3]{ay} cdot sqrt[3]{bx}} ] Simplifying, we get: [ frac{sqrt[3]{ay} + sqrt[3]{bx}}{2} geq sqrt[3]{sqrt[3]{ay cdot bx}} ] [ frac{sqrt[3]{ay} + sqrt[3]{bx}}{2} geq sqrt[3]{sqrt[3]{abxy}} ] 9. Since (x, y < 1), we have (xy < 1). Therefore: [ sqrt[3]{abxy} < sqrt[3]{ab} ] 10. From step 6, we know that: [ sqrt[3]{ab} leq 1 ] 11. Combining these results, we get: [ frac{sqrt[3]{ay} + sqrt[3]{bx}}{2} leq frac{1}{2} ] [ sqrt[3]{ay} + sqrt[3]{bx} leq 1 ] The final answer is ( boxed{ sqrt[3]{ay} + sqrt[3]{bx} leq 1 } )

answer:Let's assume the initial income is (I), the initial expenditure is (E), and the initial savings is (S). According to the given ratio of expenditure to savings, (E:S = 3:2). This means that for every 3 parts of expenditure, there are 2 parts of savings. We can express the expenditure and savings as follows: (E = 3x) (S = 2x) Where (x) is a common factor for both expenditure and savings. Since income is the sum of expenditure and savings, we have: (I = E + S) (I = 3x + 2x) (I = 5x) Now, the income increases by 15%, so the new income (I') is: (I' = I + 0.15I) (I' = 5x + 0.15(5x)) (I' = 5x(1 + 0.15)) (I' = 5x(1.15)) (I' = 5.75x) The savings increase by 6%, so the new savings (S') is: (S' = S + 0.06S) (S' = 2x + 0.06(2x)) (S' = 2x(1 + 0.06)) (S' = 2x(1.06)) (S' = 2.12x) Since the new income is the sum of the new expenditure (E') and the new savings (S'), we have: (I' = E' + S') (5.75x = E' + 2.12x) To find the new expenditure (E'), we subtract the new savings from the new income: (E' = 5.75x - 2.12x) (E' = 3.63x) Now, we need to find the percentage increase in expenditure. The initial expenditure was (3x), and the new expenditure is (3.63x). The increase in expenditure is: (Increase = E' - E) (Increase = 3.63x - 3x) (Increase = 0.63x) The percentage increase in expenditure is: (% Increase = frac{Increase}{E} times 100) (% Increase = frac{0.63x}{3x} times 100) (% Increase = frac{0.63}{3} times 100) (% Increase = 0.21 times 100) (% Increase = 21%) Therefore, the expenditure should increase by boxed{21%} .

answer:Given the curve in polar coordinates: rho = 2varphi quad text{for} quad 0 leq varphi leq frac{4}{3} 1. The length (L) of an arc defined in polar coordinates ((rho, varphi)) from (varphi = varphi_0) to (varphi = varphi_1) is given by: L = int_{varphi_0}^{varphi_1} sqrt{left(rho(varphi)right)^2 + left(frac{drho}{dvarphi}right)^2} , dvarphi 2. For the given curve, (rho = 2 varphi), we first compute (frac{drho}{dvarphi}): frac{drho}{dvarphi} = frac{d}{dvarphi}(2varphi) = 2 3. Substitute (rho = 2varphi) and (frac{drho}{dvarphi} = 2) into the formula for arc length: L = int_{0}^{frac{4}{3}} sqrt{(2 varphi)^2 + 2^2} , dvarphi 4. Simplify the integrand: L = int_{0}^{frac{4}{3}} sqrt{4varphi^2 + 4} , dvarphi = int_{0}^{frac{4}{3}} sqrt{4(varphi^2 + 1)} , dvarphi = 2 int_{0}^{frac{4}{3}} sqrt{varphi^2 + 1} , dvarphi 5. To solve the integral, we use the known formula: int sqrt{x^2 + a^2} , dx = frac{x}{2} sqrt{x^2 + a^2} + frac{a^2}{2} ln left| x + sqrt{x^2 + a^2} right| + C Here, (x = varphi) and (a = 1). 6. Apply this formula to our integral: 2 int_{0}^{frac{4}{3}} sqrt{varphi^2 + 1} , dvarphi = 2 left[ frac{varphi}{2} sqrt{varphi^2 + 1} + frac{1}{2} ln left| varphi + sqrt{varphi^2 + 1} right| right]_{0}^{frac{4}{3}} 7. Substitute the limits (varphi = 0) and (varphi = frac{4}{3}): L = left[ varphi sqrt{varphi^2 + 1} + ln left| varphi + sqrt{varphi^2 + 1} right| right]_{0}^{frac{4}{3}} L = left[ left( frac{4}{3} sqrt{left(frac{4}{3}right)^2 + 1} right) + ln left| frac{4}{3} + sqrt{left(frac{4}{3}right)^2 + 1} right| right] - left[ 0 cdot sqrt{0^2 + 1} + ln left| 0 + sqrt{0^2 + 1} right| right] 8. Evaluate inside the brackets: L = left( frac{4}{3} sqrt{frac{16}{9} + 1} + ln left| frac{4}{3} + sqrt{frac{16}{9} + 1} right| right) - (0 + ln 1) L = frac{4}{3} cdot sqrt{frac{25}{9}} + ln left| frac{4}{3} + frac{5}{3} right| 9. Further simplification yields: L = frac{4}{3} cdot frac{5}{3} + ln left| frac{4}{3} + frac{5}{3} right| L = frac{20}{9} + ln left(3right) 10. Combining the logarithm: L = frac{20}{9} + ln 3 # Conclusion: boxed{frac{20}{9} + ln 3}

answer:1. **Identify the highest and lowest prices:** Assume the highest price of milk was 22 in Month 2, and the lowest price was 16 in Month 4. 2. **Calculate the percentage increase from the lowest to the highest price:** [ text{Percentage Increase} = left(frac{text{Higher Value} - text{Lower Value}}{text{Lower Value}}right) times 100% ] Substituting the given values: [ text{Percentage Increase} = left(frac{22 - 16}{16}right) times 100% = left(frac{6}{16}right) times 100% = 37.5% ] 3. **Conclusion and verification of average price:** The average price given is 12. Verify if this aligns with the scenario: If the price fluctuates between 16 and 22, then 12 could not be the average price in this range. Hence, this part of the problem scenario has an issue. Conclusion: Ignoring the average price, the highest price is 37.5% more than the lowest price. [37.5%] is the required percentage increase. The final answer is boxed{textbf{(C)} 37.5}