To find when the containers hold the same amount of water, we need to set the two equations equal to each other and solve for x:
16x + 104 = -2x^2 + 40x + 160
Rearranging to put the equation in standard form:
2x^2 + 24x + 56 = 0
Dividing the equation by 2 to simplify:
x^2 + 12x + 28 = 0
Factoring the quadratic equation:
(x + 2)(x + 14) = 0
Setting each factor equal to zero:
x + 2 = 0
x = -2
x + 14 = 0
x = -14
Since the time after the start cannot be negative, we discard the negative solution x = -14. Therefore, the containers hold the same amount of water at x = -2 minutes, which means they have the same amount of water when they are both filled initially.
Substitute x = -2 back into either equation to find how much water they hold:
Container A: y = 16(-2) + 104 = 72 ounces
Container B: y = -2(-2)^2 + 40(-2) + 160 = 72 ounces
Therefore, both containers hold 72 ounces of water when they have the same amount.
Water is added to two containers for 16 minutes. The equations below model the ounces of water, y, in each container after x minutes. At the time after the start when the containers hold the same amount of water, how much water do they hold?
Container A: y=16x+104
Container b: y=-2x^2+40x+160
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1 answer