answer:Let's denote the length of the rectangle as ( l ) and the width as ( w ). The perimeter ( P ) of a rectangle is given by ( 2l + 2w ), and the area ( A ) as ( lw ). Given that Maria uses all her fencing material, the perimeter is 225 feet. She also wants the area to be eight times the perimeter. From the problem, we derive the equations: [ 2l + 2w = 225 ] [ lw = 8 times 225 = 1800 ] Solving for one variable in terms of the other from the perimeter equation: [ 2l + 2w = 225 implies l + w = 112.5 implies l = 112.5 - w ] Substitute ( l ) in terms of ( w ) into the area equation: [ (112.5 - w)w = 1800 ] [ 112.5w - w^2 = 1800 ] [ w^2 - 112.5w + 1800 = 0 ] Using the quadratic formula, ( w = frac{-b pm sqrt{b^2 - 4ac}}{2a} ) where ( a = 1, b = -112.5, c = 1800 ): [ w = frac{112.5 pm sqrt{112.5^2 - 4 times 1 times 1800}}{2} ] [ w = frac{112.5 pm sqrt{12656.25 - 7200}}{2} = frac{112.5 pm sqrt{5456.25}}{2} ] [ w = frac{112.5 pm 73.85}{2} ] [ w = 93.175 text{ or } w = 19.325 ] Thus, the corresponding values for ( l ) are: [ l = 112.5 - 93.175 = 19.325 text{ and } l = 112.5 - 19.325 = 93.175 ] Therefore, the largest side of the garden is ( boxed{93.175 text{ feet}} ).

answer:Consider the equation given by the problem condition: frac{2}{x} + frac{8}{y} = 1. To find the minimum value of x + y, let us first express x + y as the product of itself and the equation above: begin{align*} x+y &= (x+y) cdot 1 &= (x+y) left( frac{2}{x} + frac{8}{y} right) &= 2cdot left(frac{x}{x}right) + 8cdot left(frac{y}{y}right) + frac{8x}{y} + frac{2y}{x} &= 2 + 8 + frac{8x}{y} + frac{2y}{x}. end{align*} Now, by applying the AM-GM inequality, which states that the arithmetic mean is greater than or equal to the geometric mean for non-negative real numbers, we can compare the sum to twice the square root of the product of its non-constant terms: x + y geq 10 + 2sqrt{frac{8x}{y} cdot frac{2y}{x}} = 10 + 2sqrt{16} = 10 + 8. Thus, we find that the minimum value of x + y is 18. Therefore, we have: boxed{x + y geq 18}. This is the minimum value of x + y given the condition frac{2}{x} + frac{8}{y} = 1. The method used here is a common application of transforming the equation using the number '1', a technique often used to solve optimization problems with inequalities.

answer:First, we subtract the two numbers. Using vertical subtraction: [ begin{array}{@{}c@{;}c@{}c@{}c@{}c@{}c@{}c@{}c} & & 9 & 6. & 8 & 6 & 5 - & & 4 & 5. & 2 & 3 & 9 cline{1-7} & & 5 & 1. & 6 & 2 & 6 end{array} ] The result of the subtraction is 51.626. To round this to the nearest tenth, we consider the hundredths place, which is 2. Since 2 is less than 5, the tenths place remains as 6. The answer after rounding is therefore boxed{51.6}.

answer:To determine whether (pi(a + b)) or (2 pi sqrt{ab}) is a better approximation to the perimeter (p(a, b)) of an ellipse with semi-axes (a) and (b) when (b / a) is near 1, we analyze the expressions given. 1. **Initial Perimeter Formula Representation**: The perimeter (p(a, b)) of the ellipse can be represented by: [ p(a, b) = 4 int_0^{pi/2} sqrt{a^2 sin^2 t + b^2 cos^2 t} , dt ] 2. **Approximation Using Small Parameter (varepsilon)**: For (b = asqrt{1 - varepsilon}), where (varepsilon) is a small positive number, [ b^2 = a^2(1 - varepsilon) ] 3. **Expression Manipulation**: Substitute (b^2 = a^2(1 - varepsilon)) into the perimeter formula: [ p(a, b) = 4 int_0^{pi/2} sqrt{a^2 sin^2 t + a^2 (1 - varepsilon) cos^2 t} , dt ] Factor out (a^2): [ p(a, b) = 4a int_0^{pi/2} sqrt{sin^2 t + (1 - varepsilon) cos^2 t} , dt ] 4. **Series Expansion**: Using a Taylor series expansion for (sqrt{1 - x}) near zero (( (1 - x)^{1/2} approx 1 - frac{x}{2} - frac{x^2}{8} )): [ sqrt{sin^2 t + (1 - varepsilon) cos^2 t} = sqrt{1 - varepsilon cos^2 t} approx 1 - frac{varepsilon cos^2 t}{2} - frac{(varepsilon cos^2 t)^2}{8} ] 5. **Integral Approximation**: Perform the integral up to the second order in (varepsilon): [ p(a, b) = 4a int_0^{pi/2} left(1 - frac{varepsilon cos^2 t}{2} - frac{varepsilon^2 cos^4 t}{8}right) dt ] [ p(a, b) = 4a left( frac{pi}{2} - frac{varepsilon}{2} int_0^{pi/2} cos^2 t , dt - frac{varepsilon^2}{8} int_0^{pi/2} cos^4 t , dt right) ] Using the integrals (int_0^{pi/2} cos^2 t , dt = frac{pi}{4}) and (int_0^{pi/2} cos^4 t , dt = frac{3pi}{16}), we have: [ p(a, b) = 4a left( frac{pi}{2} - frac{varepsilon pi}{8} - frac{3 varepsilon^2 pi}{128} right) ] [ p(a, b) = 2a pi left( 1 - frac{varepsilon}{4} - frac{3 varepsilon^2}{64} right) ] 6. **Approximations (pi(a + b)) and (2 pi sqrt{ab})**: Evaluate (pi(a + b)): [ b = asqrt{1 - varepsilon} = a left(1 - frac{varepsilon}{2} - frac{varepsilon^2}{8}right) ] [ pi(a + b) = pi a left(1 + 1 - frac{varepsilon}{2} - frac{varepsilon^2}{8}right) = 2 pi a left(1 - frac{varepsilon}{4} - frac{varepsilon^2}{64}right) ] Evaluate (2 pi sqrt{ab}): [ ab = a^2 sqrt{1 - varepsilon} ] [ 2 pi sqrt{ab} = 2 pi a sqrt{1 - varepsilon} = 2 pi a left(1 - frac{varepsilon}{4} - frac{varepsilon^2}{8}right) ] 7. **Comparison of Errors**: The errors are calculated for each approximation: [ pi(a + b) = 2 pi a left(1 - frac{varepsilon}{4} - frac{varepsilon^2}{64}right) implies text{Error} = frac{pi a varepsilon^2}{32} ] [ 2 pi sqrt{ab} = 2 pi a left(1 - frac{varepsilon}{4} - frac{varepsilon^2}{8}right) implies text{Error} = frac{3 pi a varepsilon^2}{32} ] 8. **Conclusion**: Since the error in (pi(a + b)) is (frac{pi a varepsilon^2}{32}) and the error in (2 pi sqrt{ab}) is (frac{3 pi a varepsilon^2}{32}), (pi(a + b)) has a smaller error and is a better approximation for the perimeter (p(a, b)). [ boxed{pi(a + b)} ]