Question

A manufacturing company produces a product from one raw material using a process AA. This process generates total solid wastes (W_AW
A
​
) (in tonne) represented as W_A(r) = \frac{1}{15000}(-2r^{3}+10r^{2}+400r)W
A
​
(r)=
15000
1
​
(−2r
3
+10r
2
+400r), where rr is the amount of raw material used in tonne and r \in (0, 10)r∈(0,10). If the company uses a different process BB to produce the same product from same raw material, then the total solid waste generated is W_BW
B
​
(in tonne) represented as W_B(r) = \frac{1}{10000}(-2.2r^{3}+11r^{2}+440r)W
B
​
(r)=
10000
1
​
(−2.2r
3
+11r
2
+440r). The company spends ₹5,0005,000 in waste treatment using the process AA by consuming 1 tonne of raw material. How much extra amount will the company have to pay in waste treatment for consuming 1 tonne of raw material if it uses the process BB? (Enter your answer till two decimal places).

Answer 1

Answer:

Step-by-step explanation:

W  

A

​

(1)=  

15000

1

​

(−2(1)  

3

+10(1)  

2

+400(1))

=0.0272(tonne)=0.0272(tonne)

W_B(1)=\dfrac{1}{10000}(-2.2(1)^3+11(1)^2+440(1))W  

B

​

(1)=  

10000

1

​

(−2.2(1)  

3

+11(1)  

2

+440(1))

=0.04488(tonne)=0.04488(tonne)

0.04488-0.0272=0.017680.04488−0.0272=0.01768

\dfrac{0.01768}{0.04488}(5000)=1969.70  

0.04488

0.01768

​

(5000)=1969.70

Answer 2

The answer is 32172334.56.

Given:

Waste generated by process A is $W_A(r) = \frac{1}{15000}(-2r^{3}+10r^{2}+400r)$

Waste generated by process B is $W_B(r) = \frac{1}{10000}(-2.2r^{3}+11r^{2}+440r)$

r = 1 tonne

The company spends 50005000 in waste using process A.

To Find:
How much extra amount will the company have to pay in waste treatment for consuming 1 tonne of raw material if it uses the process B

Solution:

The amount the company will have to spend is directly proportional to the amount of waste generated.

Hence the amount of waste generated through process A is

$W_A(1) = \frac{1}{15000}(-2.1^{3}+10.1^{2}+400.1)\\\\W_A = \frac{408}{15000}$

Hence the amount of waste generated through process B is

$W_B(1) = \frac{1}{10000}(-2.2.1^{3}+11.1^{2}+440.1)\\\\W_B=\frac{447}{10000}$

Hence the spending of the company by using process B is

$50005000*\frac{447}{10000} *\frac{15000}{408}=82177334.56$

$82177334.56-50005000=32172334.56$

Hence the company would have to spend an extra 32172334.56.

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