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BartSMP [9]
3 years ago
5

Thermometer A shows the temperature in the morning. Thermometer B shows the temperature in the evening. What is the difference i

n the temperatures?
Mathematics
1 answer:
Evgesh-ka [11]3 years ago
3 0

Answer:

(Thermometer B reading - Thermometer A reading)

Step-by-step explanation:

The thermometer reading aren't given in the question.

However, hypothetically.

The difference between two temperature values (morning and evening values) would be :

Temperature in the evening - morning temperature

Therefore,

If ;

Thermometer A reading = morning temperature

Thermometer B reading = evening temperature

Difference in the temperature :

(Thermometer B reading - Thermometer A reading)

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A rectangular box with a volume of 684 ftcubed is to be constructed with a square base and top. The cost per square foot for the
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Answer:

The dimensions of the rectangular box is 36.23 ft×36.23 ft×4.345 ft.  

Minimum cost=2046.16 cents      

Step-by-step explanation:

Given that a rectangular box with a volume of 684 ft³.

The base and the top of the rectangular box is square in shape

Let the length and width of the rectangular box be x.

[since the base is square in shape,  length=width]

and the height of the rectangular box be h.

The volume of rectangular box is = Length ×width × height

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x^2h=684\Rightarrow h=\frac{684}{x^2}  (1)

The area of the base and top of rectangular box is = x² ft²

The surface area of the sides= 2(length+width) height

                                               =2(x+x)h

                                               =4xh ft²

The total cost to construct the rectangular box is

=[(x²×20)+(x²×15)+(4xh×1.5)] cents  

=(20x²+15x²+6xh) cents

=(25x²+6xh) cents

Total cost= C(x).

C(x) is in cents.

∴C(x)=25x²+6xh

Putting h=\frac{684}{x^2}

C(x)=25x^2+6x\times\frac{684}{x^2} \Rightarrow C(x)=25x^2+\frac{4104}{x}

Differentiating with respect to x

C'(x)=50x-\frac{4104}{x^2}

To find minimum cost, we set C'(x)=0

\therefore50x-\frac{4104}{x^2}=0\\\Rightarrow50x=\frac{4104}{x^2}\\\Rightarrow x^3=\frac{4104}{50}\Rightarrow x\approx 4.345 ft.

Putting the value x in equation (1) we get

h=\frac{684}{(4.345)^2}

 ≈36.23 ft.

The dimensions of the rectangular box is 36.23 ft×36.23 ft×4.345 ft.  

Minimum cost C(x)=[25(4.345)²+10(4.345)(36.23)] cents

                               =2046.16 cents      

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