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Salsk061 [2.6K]

3 years ago

10

Please help, need asap

1 answer:

rodikova [14]3 years ago

8 0

For the first option, since when x is going down (to the left) the function is going up, it's not approaching 0. For the second option, since when x is going up (to the right) it's going up, it's not approaching negative infinity (negative infinity is all the way down). For the third one, since when x is going down the y values are climbing, we can assume that the function's values go to positive infinity. For the last one, since when x=0 y=0, when x=0 the function does not go to infinity

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Answer:

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Step-by-step explanation:

For maximum, we use the whole length

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supposed the x length was used for the square,

the length of the side of the square = x/4m

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Area = $\frac{\sqrt{3} }{4} a^{2} = \frac{\sqrt{3} }{4} (\frac{23 - x}{3} )^{2}$

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$\frac{d^{2}A }{dx^{2} } = \frac{1}{8} + \frac{\sqrt{3} }{18} > 0$ , therefore it is minimum

$\frac{dA}{dx} = 0 \\\\$

$\frac{x}{8} - \frac{\sqrt{3} }{18} (23 - x) = 0\\$

x = 10.00m

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