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Gnesinka [82]
2 years ago
7

Maximize Q=xy^2, where x and y are positive numbers, such that x + y^2=10

Mathematics
1 answer:
mr Goodwill [35]2 years ago
8 0
X+y^2=10
minus x from both sides
y^2=10-x
sub (10-x) for y^2 in other equation
Q=x(10-x)
Q=10x-x^2
now find the maximum value
take the derititive
dQ/dx=10-2x
it is zero at x=5
below that, it is positive
after 5, it is negative
max at x=5

solve for y
y^2=10-x
y^2=10-5
y^2=5
sqrt both sides
y=√5



x=5
y=√5
the max value is 25
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Ainat [17]

Answer:

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Step-by-step explanation:

<em>Subtract by 150 from both sides of equation.</em>

<em>150-x-150=87-150</em>

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<em>87-150=-63</em>

<em>-x=-63</em>

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<em>I hope this helps you, and have a wonderful day!</em>

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3 years ago
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Sauron [17]

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<h3>How to determine the equation of the directrix?</h3>

The parabola equation is given as:

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By comparing both equations, we have:

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Hence, the directrix of the parabola is y =  \frac {-49}{16}

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<span> </span>

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