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3 years ago

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What is the range of f(x)=3^x+9

1 answer:

V125BC [204]3 years ago

7 0

$range \: = 9 \: \: \infty$

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An aircraft manufacturer wants to determine the best selling price for a new airplane. The company estimates that the initial co

Blizzard [7]

Answer:

$(a)$ $C(x)=500+20x-5x^{\frac{3}{4}}+0.01x^2$

$p(x)=320-7.7p$

$R(x)=(320-7.7p)p=320p-7.7p^2$

$(b)$ $x=82 \text{planes}$

$(c)$ $p=\$30.91 M\;\; \text{per plane}$

$(d)$ maximum profit $=\$ 15.90M$

Step-by-step explanation:

Given that,

The company estimates that the initial cost of designing the aeroplane and setting up the factories in which to build it will be 500 million dollars.

The additional cost of manufacturing each plane can be modelled by the function.

$m(x)=20x-5x^{\frac{3}{4}}+0.01x^2$

$(a)$ Find the cost, demand (or price), and revenue functions.

$C(x)=500+20x-5x^{\frac{3}{4}}+0.01x^2$

$p(x)=320-7.7p$

$R(x)=(320-7.7p)p=320p-7.7p^2$

$(b)$ Find the production level that maximizes profit.

$f=R(x)-C(x)$

$\Rightarrow f=320p-7.7p^2-(500+20x-5x^{\frac{3}{4}}+0.01x^2)$

$\Rightarrow df=320dp-15.4pdp-20dx+5(\frac{3}{4} )x^{\frac{-1}{4} }dx-0.02xdx$

$x=320-7.7p$

$p=\frac{320-x}{7.7}$

$\frac{dp}{dx} = \frac{-1}{7.7}$

$\frac{df}{dx}=\frac{320}{-7.7} -\frac{15.4(320-x) }{7.7(\frac{-1}{7.7} )}-20+5\frac{3}{4} x^{\frac{-1}{4}} -0.02x=0$

$\Rightarrow -41.5584+83.1169-0.2597x-20+3.75x^{\frac{-1}{4} }-0.02x=0$

$\Rightarrow 21.5585+3.75x^{\frac{-1}{4} }-0.279x=0$

$\Rightarrow x=82 \text{planes}$

$(c)$ Find the associated selling price of the aircraft that maximizes profit.

$p=\frac{320-82}{7.7}$

$\Rightarrow p=\$30.91 M\;\; \text{per plane}$

$(d)$ Find the maximum profit.

Manufacturing cost of one plane is:

$m(1)=20-5+0.01$

$=\$15.01 M$

maximum profit $=\$(30.91-15.01)M$

$=\$15.90M$

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I NEED HELP PLEASE, THANKS! :)

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Answer: It would be the second one.

Step-by-step explanation:

Inverse means opposite.

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