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The "break-even point" for a company is the number of units sold (other than 0 units) for which: " Profit "=" Revenue "-" Cost "=0 Production is profitable only when revenue is greater than cost. The monthly profit of a company selling x units is given by the quadratic function: P(x)=-(1)/(200)x^(2)+30 x Which of the following equivalent expressions displays the break-even point as a constant or coefficient? Choose 1 answer: (A) -(1)/(200)((x-3,000)^(2)-9:} (B) -(1)/(200)(x-3,000)^(2)+45 (C) -(1)/(-)x(x-4,000)+11

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  1. Profit function and break-even points: We are given the profit function P(x)=−1200x2+30x P(x) = -\frac{1}{200}x^2 + 30x . To find the break-even points, we need to set P(x) P(x) to zero and solve for x x .0=−1200x2+30x0 = -\frac{1}{200}x^2 + 30x
  2. Factoring out x x and solving the quadratic equation: To solve the quadratic equation, we can factor out x x :
    0=x(−1200x+30) 0 = x\left(-\frac{1}{200}x + 30\right)
  3. Finding the break-even point other than 00 units: We now have two solutions for xx when we set each factor equal to zero:\newlinex=0x = 0 or −1200x+30=0-\frac{1}{200}x + 30 = 0\newlineSince we are looking for the break-even point other than 00 units, we will solve the second equation:\newline−1200x+30=0-\frac{1}{200}x + 30 = 0
  4. Solving for x and determining the correct expression: Multiply both sides by −200-200 to get rid of the fraction:\newlinex−6000=0x - 6000 = 0
  5. Checking the options and evaluating the expressions: Add 60006000 to both sides to solve for xx:\newlinex=6000x = 6000\newlineThis is the break-even point, but we need to find which equivalent expression shows this as a constant or coefficient.
  6. Checking the options and evaluating the expressions: Add 60006000 to both sides to solve for xx:\newlinex=6000x = 6000\newlineThis is the break-even point, but we need to find which equivalent expression shows this as a constant or coefficient. Let's examine the choices given:\newline(A) −(1200)((x−3,000)2−9,000,000)-(\frac{1}{200})((x-3,000)^2 - 9,000,000)\newline(B) −(1200)(x−3,000)2+45-(\frac{1}{200})(x-3,000)^2 + 45\newline(C) −(1200)x(x−4,000)+11-(\frac{1}{200})x(x-4,000) + 11\newlineWe need to find which one of these, when set equal to zero, will give us x=6000x = 6000 as a solution.
  7. Checking the options and evaluating the expressions: Add 60006000 to both sides to solve for xx:\newlinex=6000x = 6000\newlineThis is the break-even point, but we need to find which equivalent expression shows this as a constant or coefficient. Let's examine the choices given:\newline(A) −(1200)((x−3,000)2−9,000,000)-(\frac{1}{200})((x-3,000)^2 - 9,000,000)\newline(B) −(1200)(x−3,000)2+45-(\frac{1}{200})(x-3,000)^2 + 45\newline(C) −(1200)x(x−4,000)+11-(\frac{1}{200})x(x-4,000) + 11\newlineWe need to find which one of these, when set equal to zero, will give us x=6000x = 6000 as a solution. Let's check option (A):\newline0=−(1200)((x−3,000)2−9,000,000)0 = -(\frac{1}{200})((x-3,000)^2 - 9,000,000)\newlineIf we expand this, we will not get the original quadratic equation we started with, so this cannot be the correct expression.
  8. Checking the options and evaluating the expressions: Add 60006000 to both sides to solve for xx:\newlinex=6000x = 6000\newlineThis is the break-even point, but we need to find which equivalent expression shows this as a constant or coefficient. Let's examine the choices given:\newline(A) −(1200)((x−3,000)2−9,000,000)-(\frac{1}{200})((x-3,000)^2 - 9,000,000)\newline(B) −(1200)(x−3,000)2+45-(\frac{1}{200})(x-3,000)^2 + 45\newline(C) −(1200)x(x−4,000)+11-(\frac{1}{200})x(x-4,000) + 11\newlineWe need to find which one of these, when set equal to zero, will give us x=6000x = 6000 as a solution. Let's check option (A):\newline0=−(1200)((x−3,000)2−9,000,000)0 = -(\frac{1}{200})((x-3,000)^2 - 9,000,000)\newlineIf we expand this, we will not get the original quadratic equation we started with, so this cannot be the correct expression. Let's check option (B):\newline0=−(1200)(x−3,000)2+450 = -(\frac{1}{200})(x-3,000)^2 + 45\newlineIf we multiply by −200-200 and set it equal to zero, we get:\newlinexx00\newlineThis does not give us x=6000x = 6000 as a solution, so this is not the correct expression.
  9. Checking the options and evaluating the expressions: Add 60006000 to both sides to solve for xx:\newlinex=6000x = 6000\newlineThis is the break-even point, but we need to find which equivalent expression shows this as a constant or coefficient. Let's examine the choices given:\newline(A) −(1200)((x−3,000)2−9,000,000)-(\frac{1}{200})((x-3,000)^2 - 9,000,000)\newline(B) −(1200)(x−3,000)2+45-(\frac{1}{200})(x-3,000)^2 + 45\newline(C) −(1200)x(x−4,000)+11-(\frac{1}{200})x(x-4,000) + 11\newlineWe need to find which one of these, when set equal to zero, will give us x=6000x = 6000 as a solution. Let's check option (A):\newline0=−(1200)((x−3,000)2−9,000,000)0 = -(\frac{1}{200})((x-3,000)^2 - 9,000,000)\newlineIf we expand this, we will not get the original quadratic equation we started with, so this cannot be the correct expression. Let's check option (B):\newline0=−(1200)(x−3,000)2+450 = -(\frac{1}{200})(x-3,000)^2 + 45\newlineIf we multiply by −200-200 and set it equal to zero, we get:\newline0=(x−3,000)2−9,0000 = (x-3,000)^2 - 9,000\newlineThis does not give us x=6000x = 6000 as a solution, so this is not the correct expression. Let's check option (C):\newline0=−(1200)x(x−4,000)+110 = -(\frac{1}{200})x(x-4,000) + 11\newlineIf we multiply by −200-200 and set it equal to zero, we get:\newline0=x(x−4,000)−2,2000 = x(x-4,000) - 2,200\newlineThis simplifies to:\newline0=x2−4,000x−2,2000 = x^2 - 4,000x - 2,200\newlineThis does not match our original quadratic equation, so this is not the correct expression either.
  10. Checking the options and evaluating the expressions: Add 60006000 to both sides to solve for xx:\newlinex=6000x = 6000\newlineThis is the break-even point, but we need to find which equivalent expression shows this as a constant or coefficient. Let's examine the choices given:\newline(A) −(1/200)((x−3,000)2−9,000,000)-(1/200)((x-3,000)^2 - 9,000,000)\newline(B) −(1/200)(x−3,000)2+45-(1/200)(x-3,000)^2 + 45\newline(C) −(1/200)x(x−4,000)+11-(1/200)x(x-4,000) + 11\newlineWe need to find which one of these, when set equal to zero, will give us x=6000x = 6000 as a solution. Let's check option (A):\newline0=−(1/200)((x−3,000)2−9,000,000)0 = -(1/200)((x-3,000)^2 - 9,000,000)\newlineIf we expand this, we will not get the original quadratic equation we started with, so this cannot be the correct expression. Let's check option (B):\newline0=−(1/200)(x−3,000)2+450 = -(1/200)(x-3,000)^2 + 45\newlineIf we multiply by −200-200 and set it equal to zero, we get:\newline0=(x−3,000)2−9,0000 = (x-3,000)^2 - 9,000\newlineThis does not give us x=6000x = 6000 as a solution, so this is not the correct expression. Let's check option (C):\newline0=−(1/200)x(x−4,000)+110 = -(1/200)x(x-4,000) + 11\newlineIf we multiply by −200-200 and set it equal to zero, we get:\newline0=x(x−4,000)−2,2000 = x(x-4,000) - 2,200\newlineThis simplifies to:\newline0=x2−4,000x−2,2000 = x^2 - 4,000x - 2,200\newlineThis does not match our original quadratic equation, so this is not the correct expression either. Upon reviewing the options and our calculations, it seems that none of the given choices correctly represents the break-even point as a constant or coefficient in the equivalent expression. There might be an error in the options provided or in the interpretation of the question.

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