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3 years ago

7

3) Bill weighs 220 pounds and is losing 4 pounds each month. Phil weigh 250 pounds and is losing 10 pounds each month. How many

months, m, will it take for Bill to weigh the same as Phil? How much will they weigh?

1 answer:

Anit [1.1K]3 years ago

6 0

5 months and 200 pounds

4x5 = 20

220-20

200

10x5 = 50

250-50

200

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3 years ago

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3 years ago

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<h2>THE KEY IS TO FIND OUT THE COST FUNCTION, the calculations are very easy!!!</h2><h2></h2><h3>In order to find the cost function, take a look at the drawing attached. </h3>

We can see the river (sort of) that is 3 km wide and the storage tanks on the other side of the river 8 km apart.

<h3 />

Laying pipes under (across) the river costs 1,000,000 the km & laying pipes over land costs 500,000 per km.

<h3 /><h3>So basically the cost function is 1,000,000 multiplied by something plus 500,000 multiplied by another something.</h3><h3 />

The distance across the river can be found by using Pythagoras Theorem. A side is 3 km the other is unknown, so we call it X. And it is equal to:

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And we multiply it by 1,000,000; the cost of laying pipe under the river, the we get:

$1000000\sqrt{9+x^{2}$

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So the cost function f(x) would be:

$f(x)=1000000\sqrt{9+x^{2}} + 500000(8-x)$

<h2>From here, we just have to differentiate and the derivative found must be equal to zero in order to minimize cost. </h2><h3>The value of x when the derivative is zero is plugged in the original function to get the cost.</h3><h3 /><h2>LET'S DO THIS</h2>

$f(x)=1000000\sqrt{9+x^{2}} + 500000(8-x)\\f(x)=1000000(9+x^{2})^{1/2}+4000000-500000x\\f'(x)=\frac{1}{2} 1000000(9+x^{2})^{-1/2}(2x)-500000\\\\f'(x)=\frac{1000000x}{\sqrt{9+x^2}} - 500000$

<h2>f'(x)=0</h2>

$f'(x)=\frac{1000000x}{\sqrt{9+x^2}} - 500000=0\\\frac{1000000x}{\sqrt{9+x^2}} = 500000\\\frac{2x}{\sqrt{9+x^2}} = 1\\2x={\sqrt{9+x^2}}\\4x^2=9+x^2\\3x^2=9\\x^2=3\\x=\sqrt{3} \\$

And we plug square root of 3 in the original cost function ad we get

$f(\sqrt{3} )=1000000\sqrt{9+x^{2}} + 500000(8-x)\\f(\sqrt{3})=1000000\sqrt{9+(\sqrt{3} )^{2}} + 500000(8-(\sqrt{3}))\\f(\sqrt{3})=1000000\sqrt{9+3} + 500000(8-(\sqrt{3}))\\f(\sqrt{3})=1000000\sqrt{12}+500000(6.27)\\f(\sqrt{3})=1000000(3.46)+500000(6.27)\\f(\sqrt{3})=3464101.62+3133974.60\\f(\sqrt{3})=6598076.21\\$

<h2>so the minimal cost is $6,598,076.21</h2><h2 /><h3 />

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4 years ago