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Find the value of x that maximizes y = 20x – x Find the value of x that minimizes y = x Given TC = x Given TR = 100x – x A firm's profit function is π = 120x – x

Mathematics

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41. Find the value of xx that maximizes y=20x−x2y = 20x - x^2.

  1. To find the maximum value, we take the first derivative of the function yy with respect to xx.

    dydx=ddx(20x−x2)=20−2x\frac{dy}{dx} = \frac{d}{dx}(20x - x^2) = 20 - 2x
  2. Set the first derivative equal to zero and solve for xx.

    20−2x=020 - 2x = 0 2x=202x = 20 x=10x = 10

    The value of xx that maximizes yy is 10\boxed{\text{10}}.

    42. Find the value of xx that minimizes y=x2+4x+7y = x^2 + 4x + 7.

  3. To find the minimum value, we take the first derivative of the function yy with respect to xx.

    dydx=ddx(x2+4x+7)=2x+4\frac{dy}{dx} = \frac{d}{dx}(x^2 + 4x + 7) = 2x + 4
  4. Set the first derivative equal to zero and solve for xx.

    2x+4=02x + 4 = 0 2x=−42x = -4 x=−2x = -2

    The value of xx that minimizes yy is -2\boxed{\text{-2}}.

    43. Given TC=x2+10x+50TC = x^2 + 10x + 50, find the Marginal Cost.

  5. Marginal Cost (MC) is the derivative of the Total Cost (TC) function with respect to xx.

    MC=d(TC)dxMC = \frac{d(TC)}{dx}
  6. Differentiate the given TC function.

    MC=ddx(x2+10x+50)=2x+10MC = \frac{d}{dx}(x^2 + 10x + 50) = 2x + 10

    The Marginal Cost is 2x+10\boxed{2x + 10}.

    44. Given TR=100x−x2TR = 100x - x^2, find the Marginal Revenue.

  7. Marginal Revenue (MR) is the derivative of the Total Revenue (TR) function with respect to xx.

    MR=d(TR)dxMR = \frac{d(TR)}{dx}
  8. Differentiate the given TR function.

    MR=ddx(100x−x2)=100−2xMR = \frac{d}{dx}(100x - x^2) = 100 - 2x

    The Marginal Revenue is 100−2x\boxed{100 - 2x}.

    45. A firm's profit function is π=120x−x2−200\pi = 120x - x^2 - 200. Determine the output level that maximizes profit.

  9. To maximize profit, we take the first derivative of the profit function π\pi with respect to xx.

    dπdx=ddx(120x−x2−200)=120−2x\frac{d\pi}{dx} = \frac{d}{dx}(120x - x^2 - 200) = 120 - 2x
  10. Set the first derivative equal to zero and solve for xx.

    120−2x=0120 - 2x = 0 2x=1202x = 120 x=60x = 60

    The output level that maximizes profit is 60\boxed{\text{60}}.

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