Appearance
question:As a local resident in Freeport, Texas, you are excited about the new industrial development project that promises to bring more businesses to the area. The city plans to allocate a rectangular plot of land for a new manufacturing plant. The length of the plot is 10 meters longer than its width.1. If the area of the plot is 1200 square meters, set up a quadratic equation to find the dimensions of the plot and solve for the width.2. The city also plans to build a road around the plot to facilitate transportation. If the width of the road is 5 meters, what is the new area occupied by both the plot and the road? Express the expanded area as a polynomial in terms of the width of the plot.
answer:To solve the given problem, we will proceed step-by-step. Part 1: Find the dimensions of the plot1. Set up the equation: Let the width of the plot be ( w ) meters. According to the problem, the length is 10 meters longer than the width. Therefore, the length can be expressed as ( w + 10 ) meters. The area of the rectangle is given by the formula: [ text{Area} = text{Length} times text{Width} = (w + 10) times w ] We know the area is 1200 square meters, so we set up the equation: [ w(w + 10) = 1200 ] Simplifying, we obtain the quadratic equation: [ w^2 + 10w - 1200 = 0 ]2. Solve the quadratic equation: To solve the quadratic equation ( w^2 + 10w - 1200 = 0 ), we can use the quadratic formula: [ w = frac{-b pm sqrt{b^2 - 4ac}}{2a} ] where ( a = 1 ), ( b = 10 ), and ( c = -1200 ). Calculate the discriminant: [ b^2 - 4ac = 10^2 - 4 times 1 times (-1200) = 100 + 4800 = 4900 ] The square root of 4900 is 70, so the solutions are: [ w = frac{-10 pm 70}{2} ] This gives us two possible solutions for ( w ): [ w = frac{-10 + 70}{2} = frac{60}{2} = 30 ] [ w = frac{-10 - 70}{2} = frac{-80}{2} = -40 ] Since width cannot be negative, we discard ( w = -40 ). Thus, the width is ( w = 30 ) meters. Part 2: Find the expanded area with the road3. Calculate the dimensions including the road: The road is 5 meters wide, so it adds 5 meters to each side of the plot's width and length. Therefore, the new width and length are: [ text{New Width} = w + 2 times 5 = w + 10 ] [ text{New Length} = (w + 10) + 2 times 5 = w + 20 ]4. Express the new area as a polynomial: The new area, including the road, is given by: [ text{New Area} = (text{New Width}) times (text{New Length}) = (w + 10)(w + 20) ] Expanding this expression: [ (w + 10)(w + 20) = w^2 + 20w + 10w + 200 = w^2 + 30w + 200 ] Substitute ( w = 30 ) to find the numerical value of the expanded area: [ (30)^2 + 30(30) + 200 = 900 + 900 + 200 = 2000 ]Thus, the new area occupied by both the plot and the road is 2000 square meters.Final Answer: The final answer is ( text{Width of the plot} = 30 text{ meters}, text{Expanded area} = w^2 + 30w + 200 text{ square meters} ). I hope it is correct.
question:A non-profit organization focused on enhancing community resilience in coastal regions is designing a rectangular garden to promote sustainable agricultural practices. The length of the garden is 3 meters more than twice its width. The organization wants the area of the garden to be 150 square meters in order to maximize the space for planting a variety of crops.1. Write an equation to represent the relationship between the length and the width of the garden and solve for the width.2. The organization also plans to install a circular water tank next to the garden. The cost of the tank is directly proportional to the square of its radius. If the radius of the tank is increased by 2 meters, the cost increases by 200. Find the original radius of the tank if the proportionality constant is 10 per square meter.
answer:To solve the problem, we'll tackle each part of the problem separately. Part 1: Finding the Width of the Garden1. Define the Variables: - Let ( w ) be the width of the garden in meters. - The length of the garden ( l ) is given as ( l = 2w + 3 ).2. Area Equation: - The area of the garden is given by the formula ( text{Area} = text{length} times text{width} ). - We know the area is 150 square meters, so: [ (2w + 3) times w = 150 ]3. Solve for the Width: - Expand the equation: [ 2w^2 + 3w = 150 ] - Rearrange it to form a standard quadratic equation: [ 2w^2 + 3w - 150 = 0 ]4. Use the Quadratic Formula: - The quadratic formula is ( w = frac{-b pm sqrt{b^2 - 4ac}}{2a} ), where ( a = 2 ), ( b = 3 ), and ( c = -150 ). - Calculate the discriminant: [ b^2 - 4ac = 3^2 - 4 times 2 times (-150) = 9 + 1200 = 1209 ] - Solve for ( w ): [ w = frac{-3 pm sqrt{1209}}{4} ]5. Approximate the Roots: - Calculate ( sqrt{1209} approx 34.78 ). - Therefore, the possible values for ( w ) are: [ w = frac{-3 + 34.78}{4} quad text{and} quad w = frac{-3 - 34.78}{4} ] - These simplify to: [ w approx frac{31.78}{4} approx 7.945 quad text{and} quad w approx frac{-37.78}{4} approx -9.445 ] - Since a negative width does not make sense, ( w approx 7.945 ). Part 2: Finding the Original Radius of the Water Tank1. Define the Variables: - Let ( r ) be the original radius of the tank in meters. - The cost of the tank is proportional to the square of its radius, meaning the cost ( C = k times r^2 ) where ( k = 10 ).2. Cost Increase Equation: - If the radius increases by 2 meters, the new radius is ( r + 2 ). - The cost difference due to the increase in radius is 200: [ 10 times (r + 2)^2 - 10 times r^2 = 200 ] - Simplify the equation: [ 10 times ((r + 2)^2 - r^2) = 200 ] - Divide both sides by 10: [ (r + 2)^2 - r^2 = 20 ] - Expand: [ (r^2 + 4r + 4) - r^2 = 20 ] - Simplify: [ 4r + 4 = 20 ] - Solve for ( r ): [ 4r = 16 quad Rightarrow quad r = 4 ]Final Answer: The width of the garden is approximately 7.945 meters, and the original radius of the water tank is 4 meters. I hope it is correct.
question:A tenure-track faculty member in Indigenous history at a Canadian University is preparing a lecture series about the impact of early trade networks among Indigenous peoples. She is creating a model to represent the growth of trade networks over time. She finds that the number of trade connections ( C(t) ) between different Indigenous groups can be modeled by the quadratic equation ( C(t) = t^2 + 6t + 9 ), where ( t ) represents time in decades since the beginning of the trade network.1. Determine the time ( t ) at which the number of trade connections reaches its minimum, and calculate the minimum number of trade connections.2. Additionally, the faculty member is considering the influence of external trade networks. A linear model ( E(t) = 5t + 15 ) represents the number of external trade connections over the same period. For what range of ( t ) is the number of internal trade connections ( C(t) ) greater than the number of external trade connections ( E(t) )?
answer:To solve the given problem, we need to address both parts of the question. Part 1: Determine the time ( t ) at which the number of trade connections ( C(t) ) reaches its minimum, and calculate the minimum number of trade connections.The function ( C(t) = t^2 + 6t + 9 ) is a quadratic function in the standard form ( ax^2 + bx + c ), where ( a = 1 ), ( b = 6 ), and ( c = 9 ).For a quadratic function ( ax^2 + bx + c ), the vertex (which gives the minimum value when ( a > 0 )) is located at the point ( t = -frac{b}{2a} ).1. Calculate the time ( t ) at which ( C(t) ) is minimized: [ t = -frac{b}{2a} = -frac{6}{2 times 1} = -frac{6}{2} = -3 ]2. Calculate the minimum number of trade connections by substituting ( t = -3 ) into the function ( C(t) ): [ C(-3) = (-3)^2 + 6(-3) + 9 = 9 - 18 + 9 = 0 ]Thus, the minimum number of trade connections is 0, and it occurs at ( t = -3 ). Part 2: Determine the range of ( t ) for which ( C(t) > E(t) ).Given:- ( C(t) = t^2 + 6t + 9 )- ( E(t) = 5t + 15 )We need to find when ( C(t) > E(t) ):[t^2 + 6t + 9 > 5t + 15]Subtract ( E(t) ) from both sides:[t^2 + 6t + 9 - 5t - 15 > 0][t^2 + t - 6 > 0]Now, solve the inequality ( t^2 + t - 6 > 0 ).1. Factor the quadratic equation: [ t^2 + t - 6 = (t + 3)(t - 2) ]2. Determine the critical points where the expression equals zero: [ t + 3 = 0 quad Rightarrow quad t = -3 ] [ t - 2 = 0 quad Rightarrow quad t = 2 ]These critical points divide the real number line into intervals: ( (-infty, -3) ), ( (-3, 2) ), and ( (2, infty) ).3. Test each interval to see where the inequality is satisfied: - For ( t in (-infty, -3) ): Choose ( t = -4 ), ((t + 3)(t - 2) = (-1)(-6) = 6 > 0). - For ( t in (-3, 2) ): Choose ( t = 0 ), ((t + 3)(t - 2) = (3)(-2) = -6 < 0). - For ( t in (2, infty) ): Choose ( t = 3 ), ((t + 3)(t - 2) = (6)(1) = 6 > 0).Therefore, ( C(t) > E(t) ) for ( t in (-infty, -3) cup (2, infty) ).Final Answer: The minimum number of trade connections is 0 at ( t = -3 ). The number of internal trade connections ( C(t) ) is greater than the number of external trade connections ( E(t) ) for ( t in (-infty, -3) cup (2, infty) ). I hope it is correct.
question:A marketing strategist is analyzing the sales potential of a new line of functional accessories designed for a niche market of outdoor enthusiasts. The strategist has determined that the expected monthly revenue ( R(x) ) in thousands of dollars can be modeled by the quadratic equation ( R(x) = -2x^2 + 24x - 50 ), where ( x ) is the number of thousands of units sold.1. Determine the number of units ( x ) that need to be sold in order to maximize the expected monthly revenue. What is the maximum revenue?2. To ensure profitability, the strategist has set a goal that the revenue must be at least 80,000. Find the range of unit sales ( x ) that would meet this revenue goal.
answer:To solve this problem, we need to analyze the quadratic revenue function ( R(x) = -2x^2 + 24x - 50 ). 1. Maximizing the Expected Monthly RevenueThe given function is a quadratic equation in the form ( R(x) = ax^2 + bx + c ), where ( a = -2 ), ( b = 24 ), and ( c = -50 ).The vertex form of a quadratic equation ( ax^2 + bx + c ) gives the maximum (or minimum) point of the parabola, which is also the maximum revenue in this case because ( a < 0 ) (the parabola opens downwards).The vertex ( x )-coordinate can be found using the formula:[x = -frac{b}{2a}]Substituting the values of ( a ) and ( b ):[x = -frac{24}{2 times (-2)} = frac{24}{4} = 6]Thus, the number of units ( x ) that need to be sold to maximize revenue is 6,000 units.To find the maximum revenue, substitute ( x = 6 ) back into the revenue equation:[R(6) = -2(6)^2 + 24(6) - 50][R(6) = -2(36) + 144 - 50][R(6) = -72 + 144 - 50][R(6) = 22]Since the revenue is modeled in thousands of dollars, the maximum revenue is 22,000. 2. Revenue Goal of At Least 80,000We need to find the values of ( x ) such that ( R(x) geq 80 ) (since the revenue function is in thousands of dollars, 80,000 corresponds to 80).Set the inequality:[-2x^2 + 24x - 50 geq 80]Rearrange to form a standard quadratic inequality:[-2x^2 + 24x - 130 geq 0]Divide the entire inequality by (-2) to make it easier to solve:[x^2 - 12x + 65 leq 0]To solve this inequality, first find the roots of the equation ( x^2 - 12x + 65 = 0 ) using the quadratic formula:[x = frac{-b pm sqrt{b^2 - 4ac}}{2a}]where ( a = 1 ), ( b = -12 ), and ( c = 65 ).Calculate the discriminant:[b^2 - 4ac = (-12)^2 - 4 times 1 times 65 = 144 - 260 = -116]Since the discriminant is negative, the quadratic ( x^2 - 12x + 65 = 0 ) has no real roots, implying that ( x^2 - 12x + 65 ) is always positive for all ( x ). Therefore, there is no real ( x ) that satisfies ( -2x^2 + 24x - 130 geq 0 ).This means the revenue goal of 80,000 cannot be reached with any number of units sold within the constraints of this model.Final Answer: The number of units to maximize revenue is 6,000, with a maximum revenue of 22,000. The revenue goal of 80,000 cannot be reached. I hope it is correct.
question:The choir director at a local church is organizing a special performance featuring the children's choir. The director wants to form a triangle formation with the children when they sing. The number of children in each row of the triangle increases in a linear fashion, starting with 3 children in the first row and increasing by 2 children in each subsequent row.1. If the total number of children in the choir is expressed as a quadratic equation ( n = ar^2 + br + c ), where ( r ) is the number of rows, determine the values of ( a ), ( b ), and ( c ) based on the given information.2. The director realizes they have 45 children available for the performance. Use the quadratic equation from part 1 to determine how many complete rows can be formed in this triangle configuration.
answer:To solve this problem, we need to determine the number of children in each row and then construct a quadratic equation for the total number of children based on the number of rows. Part 1: Constructing the Quadratic Equation1. Determine the number of children in each row: - The number of children in the first row is 3. - The number of children in each subsequent row increases by 2. The number of children in the ( r )-th row can be expressed as: [ text{Children in } rtext{-th row} = 3 + 2(r - 1) = 2r + 1. ]2. Sum the number of children up to ( r ) rows: The total number of children ( n ) in ( r ) rows is the sum of an arithmetic series: [ n = sum_{k=1}^{r} (2k + 1). ] Simplifying, we have: [ n = (2 times 1 + 1) + (2 times 2 + 1) + cdots + (2r + 1). ] This simplifies to: [ n = sum_{k=1}^{r} (2k + 1) = 2sum_{k=1}^{r} k + sum_{k=1}^{r} 1. ] Using the formula for the sum of the first ( r ) natural numbers, (sum_{k=1}^{r} k = frac{r(r+1)}{2}), we get: [ n = 2 left(frac{r(r+1)}{2}right) + r = r(r+1) + r = r^2 + 2r. ] Thus, the quadratic equation for the total number of children is: [ n = r^2 + 2r. ] Comparing with ( n = ar^2 + br + c ), we have: - ( a = 1 ), - ( b = 2 ), - ( c = 0 ). Part 2: Solving for the Number of Rows with 45 ChildrenWe know that:[n = r^2 + 2r.]Given ( n = 45 ), we have the equation:[r^2 + 2r = 45.]Rearranging gives us:[r^2 + 2r - 45 = 0.]This is a quadratic equation, which we can solve using the quadratic formula:[r = frac{-b pm sqrt{b^2 - 4ac}}{2a},]where ( a = 1 ), ( b = 2 ), ( c = -45 ).Calculating the discriminant:[b^2 - 4ac = 2^2 - 4 times 1 times (-45) = 4 + 180 = 184.]Thus, the solutions for ( r ) are:[r = frac{-2 pm sqrt{184}}{2}.]Simplifying:[r = frac{-2 pm sqrt{184}}{2}.]Since ( sqrt{184} ) simplifies to ( sqrt{4 times 46} = 2sqrt{46} ), we have:[r = frac{-2 pm 2sqrt{46}}{2} = -1 pm sqrt{46}.]Since ( r ) must be a positive integer, we take:[r = -1 + sqrt{46}.]Approximating ( sqrt{46} ) gives ( sqrt{46} approx 6.78 ), so:[r approx -1 + 6.78 = 5.78.]The maximum integer value ( r ) can take is 5.Final Answer: The final answer is 5. I hope it is correct.
question:As a community organizer in Cumberland County, North Carolina, you're planning a series of meetings to address local civic issues. You have two types of meeting spaces available: community halls and local libraries. 1. The cost to rent a community hall is 150 per hour, and the cost to rent a room in a library is 100 per hour. You need a total of 20 hours of meeting time. Your budget for renting these spaces is 2,400. Formulate a system of linear equations to determine how many hours you should rent each type of space. Solve for the number of hours each space should be rented.2. In addition to renting spaces, you want to ensure maximum participation by choosing optimal times for meetings. Based on previous events, the function ( P(t) = -2t^2 + 24t - 40 ) models the expected number of participants at meeting time ( t ) (in hours from 0 to 12). Determine the time ( t ) that will maximize participation, and calculate the maximum expected number of participants.
answer:To solve the given problem, we will break it down into the two parts as specified. Part 1: Renting Meeting SpacesWe need to determine how many hours to rent community halls and library rooms given the constraints on total hours and budget. Let's define:- ( x ) as the number of hours to rent a community hall.- ( y ) as the number of hours to rent a library room.From the problem, we have the following constraints:1. The total hours of meeting time needed is 20: [ x + y = 20 ]2. The budget for renting these spaces is 2,400: [ 150x + 100y = 2400 ]We now have a system of linear equations:[begin{align*}1. & quad x + y = 20 2. & quad 150x + 100y = 2400end{align*}]Let's solve this system using substitution or elimination.First, solve the first equation for ( y ):[y = 20 - x]Substitute ( y = 20 - x ) into the second equation:[150x + 100(20 - x) = 2400]Simplify and solve for ( x ):[150x + 2000 - 100x = 2400 50x + 2000 = 2400 50x = 400 x = 8]Now substitute ( x = 8 ) back into the equation ( y = 20 - x ):[y = 20 - 8 = 12]So, the solution for the number of hours is:- Rent the community hall for 8 hours.- Rent the library room for 12 hours. Part 2: Maximizing ParticipationThe function modeling expected participation is:[P(t) = -2t^2 + 24t - 40]This is a quadratic function in the form ( P(t) = at^2 + bt + c ) with ( a = -2 ), ( b = 24 ), and ( c = -40 ).To find the time ( t ) that maximizes participation, we need to find the vertex of the parabola. The vertex ( t ) for a quadratic function ( at^2 + bt + c ) occurs at:[t = -frac{b}{2a}]Substitute ( a = -2 ) and ( b = 24 ) into the formula:[t = -frac{24}{2 times -2} = frac{24}{4} = 6]Now, calculate the maximum expected number of participants by substituting ( t = 6 ) into the function ( P(t) ):[P(6) = -2(6)^2 + 24(6) - 40 = -2(36) + 144 - 40 = -72 + 144 - 40 = 32]Therefore, the time that maximizes participation is 6 hours, and the maximum expected number of participants is 32.Final Answer: The final answer is renting the community hall for 8 hours and the library for 12 hours, with maximum participation at 6 hours yielding 32 participants. I hope it is correct.