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Maximize/Minimize Quadratic Functions Help (page 4)

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By — McGraw-Hill Professional
Updated on Oct 4, 2011

Solutions

  1. We only need to find h .

     

    Quadratic Functions Solutions

    Minimize the average cost by producing 500 units.

  2. k answers the first question, h answers the second.

    Quadratic Functions Solutions

    The frog reached a maximum height of 10 inches and had traveled 12 inches horizontally when it reached its maximum height.

  3. The formula s ( t ) = − 16 t 2 + v 0 t + s 0 is in feet and seconds, so we need to convert 108 miles per hour to feet per second. There are 5280 feet in a mile and 60(60) = 3600 seconds in an hour.

    Quadratic Functions Solutions

    Replacing v 0 with 158.4 and s 0 with 10, we have the function giving the height of the projectile after t seconds, s ( t ) = − 16 t 2 + 158.4 t + 10.

    Quadratic Functions Solutions

    The projectile reaches a maximum height of 402.04 feet after 4.95 seconds.

  4. We will let x represent the number of $0.25 decreases in the ticket price. The ticket price is 12 − 0.25 x and the average number attending the games is 1000 + 25 x . Ticket revenue is (12 − 0.25 x )(1000 + 25 x ). Revenue from concession sales is 2(1000 + 25 x ). Total revenue is

    Quadratic Functions Solutions

    The ticket price that will maximize revenue is 12 − 0.25(8) = $10 and the maximum revenue is $14,400.

  5. The total area is A = L W . Because there is 1600 feet of fencing available, 2 L + 4 W = 1600. Solving this equation for L , we have L = 800 − 2 W . Substitute 800 − 2 W for L in A = LW .

    Quadratic Functions Solutions

    Maximize the enclosed area by letting the width be 200 feet and the length be 800 − 2(200) = 400 feet.

  6. The enclosed area is A = LW . Because 900 feet of fencing is available, 2 W + L = 900. Solving this for L , we have L = 900 − 2 W . We will substitute 900 − 2 W for L in A = LW .

    A = LW

    = (900 − 2 W ) W = −2 W 2 + 900 W

    Quadratic Functions Solutions

    Maximize the enclosed area by letting the width be 225 feet and the length 900 − 2(225) = 450 feet. The maximum enclosed area is 101,250 square feet.

  7. The area of the rectangle is 2 rL (the width is twice the radius of the semicircle). The area of the semicircle is half the area of a circle with radius r , Quadratic Functions Solutions . The total area of the pool is Quadratic Functions Solutions . After finding an equation for the perimeter, we will solve the equation for L and substitute this for L in Quadratic Functions Solutions . The perimeter of the rectangular part is L + 2 r + L = 2 r + 2 L . The length around the semicircle is half the circumference of a circle with radius r , Quadratic Functions Solutions . The total length around the pool is 2 L + 2 r + π r which equals 120 feet.

    Quadratic Functions Solutions

    Quadratic Functions Solutions

    Maximize the area by letting the radius of the semicircle be about 16.8 feet and the length of the rectangle about Quadratic Functions Solutions feet. The maximum area is about 1008.2 square feet.

  8. The area is A = LW . The length of the rectangle is y (the distance from (0,0) and (0, y )). The width is x (the distance from (0, 0) and ( x , 0)). The area is now A = xy . Because y = −3 x + 4, we can substitute − 3 x + 4 for y in A = xy .

    Quadratic Functions Solutions

    The maximum area is Quadratic Functions Solutions square units.

Practice problems for this concept can be found at: Quadratic Functions Practice Test.

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