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MaRussiya [10]
3 years ago
13

A 40% antifreeze solution is to be mixed with a 70% antifreeze solution to get 240 liters of a 64% solution. How many liters of

the 40% and how many liters of the 70% solutions will be used?
Mathematics
1 answer:
Alexxandr [17]3 years ago
6 0

Answer:

We have 40% antifreeze and 70% antifreeze and we need to make 240 liters of 64% antifreeze.

We set up 2 equations where "f" is the 40 % and "s" is the 70%

A) f + s = 240

B) .40f + .70s = (.64 * 240) (or 153.6)

To solve both equations we multiply A) by -.40

A) -.40 f  - .40 s = -96.00   then we add this to B)

B) .40f + .70s = 153.60

.30s = 57.6

s = 192

f = 48

48 liters of 40% = 19.20 liters of antifreeze

192 liters of  70% = 134.40 liters of antifreeze

That equals 153.60 liters of antifreeze in a TOTAL liquid amount of 240 liters.

Double Check

153.60 / 240 = 64% antifreeze.

Step-by-step explanation:

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A solid is formed by adjoining two hemispheres to the ends of a right circular cylinder. An industrial tank of this shape must h
mestny [16]

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Radius =6.518 feet

Height = 26.074 feet

Step-by-step explanation:

The Volume of the Solid formed  = Volume of the two Hemisphere + Volume of the Cylinder

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Volume of a Cylinder =\pi r^2 h

Therefore:

The Volume of the Solid formed

=2(\frac{2}{3}\pi r^3)+\pi r^2 h\\\frac{4}{3}\pi r^3+\pi r^2 h=4640\\\pi r^2(\frac{4r}{3}+ h)=4640\\\frac{4r}{3}+ h =\frac{4640}{\pi r^2} \\h=\frac{4640}{\pi r^2}-\frac{4r}{3}

Area of the Hemisphere =2\pi r^2

Curved Surface Area of the Cylinder =2\pi rh

Total Surface Area=

2\pi r^2+2\pi r^2+2\pi rh\\=4\pi r^2+2\pi rh

Cost of the Hemispherical Ends  = 2 X  Cost of the surface area of the sides.

Therefore total Cost, C

=2(4\pi r^2)+2\pi rh\\C=8\pi r^2+2\pi rh

Recall: h=\frac{4640}{\pi r^2}-\frac{4r}{3}

Therefore:

C=8\pi r^2+2\pi r(\frac{4640}{\pi r^2}-\frac{4r}{3})\\C=8\pi r^2+\frac{9280}{r}-\frac{8\pi r^2}{3}\\C=\frac{9280}{r}+\frac{24\pi r^2-8\pi r^2}{3}\\C=\frac{9280}{r}+\frac{16\pi r^2}{3}\\C=\frac{27840+16\pi r^3}{3r}

The minimum cost occurs at the point where the derivative equals zero.

C^{'}=\frac{-27840+32\pi r^3}{3r^2}

When \:C^{'}=0

-27840+32\pi r^3=0\\27840=32\pi r^3\\r^3=27840 \div 32\pi=276.9296\\r=\sqrt[3]{276.9296} =6.518

Recall:

h=\frac{4640}{\pi r^2}-\frac{4r}{3}\\h=\frac{4640}{\pi*6.518^2}-\frac{4*6.518}{3}\\h=26.074 feet

Therefore, the dimensions that will minimize the cost are:

Radius =6.518 feet

Height = 26.074 feet

5 0
2 years ago
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