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Ask QuestionPosted by Tanushri Bhayani 1 year, 8 months ago
- 2 answers
Tanushri Bhayani 1 year, 8 months ago
Posted by Tanushri Bhayani 1 year, 8 months ago
- 2 answers
Preeti Dabral 1 year, 8 months ago
Let the number of guests be x
If x > 100, then the amount received by the company will be-
{4800 - 200/10(x - 100)}x
Now, we need to maximize this amount so let us differentiate with respect to x as follows-
P = 4800x - 20(x - 100)x
P = 4800x - 20x² + 2000x
P = 6800x - 20x²
dP/dx = 6800 - 40x
differentiating again-
d²P/dx² = -40
As d²P/dx² > 0, the maximum value will occur at dP/dx = 0
⇒ 6800 - 40x = 0
40x = 6800
x = 6800/40
x = 170
Now, the amount received will be = 6800(170) - 20(170)²
= 1156000 - 578000
= 578000
Thus, the maximum amount that the company will receive is 5,78,000 when the number of guests are 170.
Posted by Tanushri Bhayani 1 year, 8 months ago
- 2 answers
Preeti Dabral 1 year, 8 months ago
Distance covered by a boat in 5 hours = 36 km
Rate upstream of boat {tex}=\frac{36}{5}{/tex} 7.2 kmph
Speed of the stream = 2.4 kmph
{tex}\therefore{/tex} Speed of the boat in still water
= (7.2 + 2.4) kmph
= 9.6 kmph
{tex}\therefore{/tex} Rate downstream of the boat
= (9.6 + 2.4) kmph
= 12 kmph
{tex}\therefore{/tex} Time taken in covering 78 km distance
{tex}=\frac{78}{12}{/tex}
= 6.5 hours or 6 hour 30 minutes.
Posted by Tanushri Bhayani 1 year, 8 months ago
- 2 answers
Preeti Dabral 1 year, 8 months ago
Let P be the principal at any time t. According to the given problem, {tex}\frac{d p}{d t}=\left(\frac{5}{100}\right) \times \mathrm{P}{/tex}
or {tex}\frac{d p}{d t}=\frac{\mathrm{P}}{20}{/tex} ...(i)
separating the variables in equation (i), we get
{tex}\frac{d p}{\mathrm{P}}=\frac{d t}{20}{/tex} ...(ii)
Integrating both sides of equation (ii), we get
log P = {tex}\frac{t}{20}{/tex} + C1
or P = {tex}e^{\frac{t}{20}} \cdot e^{\mathrm{C}_{1}}{/tex}
or P = {tex}\mathrm{C} e^{\frac{t}{20}}{/tex} (where eC1 = C) ...(iii)
Now, P = 1000, when t = 0
Substituting the values of P and t in (iii), we get C = 1000. Therefore, equation (iii), gives
P = 1000 et/20
Let t years be the time required to double the principal. Then 2000 = 1000 et/20 {tex}\Rightarrow{/tex} t = 20 loge2.
Posted by Praagna Koya 1 year, 9 months ago
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Tanushri Bhayani 1 year, 8 months ago
Posted by Ri S 1 year, 10 months ago
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Tanushri Bhayani 1 year, 8 months ago
Posted by Harrish Kumaran 1 year, 11 months ago
- 1 answers
Posted by Arpit Agrawal 1 year, 11 months ago
- 1 answers
Preeti Dabral 1 year, 11 months ago
4 - 120
= 48
62 ÷ 11
= 5
48 + 5
= 53
therefore, the test compared to the other volunteer
= 53
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- 0 answers
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- 0 answers
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- 2 answers
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- 4 answers
Tanushri Bhayani 1 year, 8 months ago
Tanushri Bhayani 1 year, 8 months ago
Tanushri Bhayani 1 year, 8 months ago
Tanushri Bhayani 1 year, 8 months ago
Posted by Manisha Behera 1 year, 11 months ago
- 1 answers
Preeti Dabral 1 year, 11 months ago
We have
f (x) = 2x3 – 6x2 + 6x + 5
or f ′(x) = 6x2 – 12x + 6 = 6 (x – 1)2
Now, f ′(x) = 0
{tex}\Rightarrow{/tex} x = 1
Thus, x = 1 is the only critical point of f . We shall now examine this point for local maxima and/or local minima of f. Observe that f ′(x) {tex}\ge{/tex} 0, for all x {tex}\in{/tex} R and in particular f ′(x) > 0, for values close to 1, to the left and to the right of 1. Therefore, by first derivative test, the point x = 1 is neither a point of local maxima nor a point of local minima. Hence x = 1 is a point of inflexion.
Posted by Deep Chakraborty 2 years, 2 months ago
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Posted by Pravee Na 1 year, 11 months ago
- 1 answers
Preeti Dabral 1 year, 11 months ago
Speed of the man in still water =8 kmph.
Speed of the river =2 kmph
Downstream =8+2=10 kmph
Upstream =8−2=6 kmph
{tex}\Rightarrow \frac{x}{10}+\frac{x}{6}=\frac{48}{60}{/tex}
⇒8x=24
⇒x=3 km
Posted by Moorthi Maths 1 year, 11 months ago
- 1 answers
Preeti Dabral 1 year, 11 months ago
Cost of raw material = x²
Cost of transportation = 2x
Property tax = 5,000
Therefore ,
(i) C(x) = x² + 2x + 5000
(ii) MC = 2x + 2
Now, x = 21,
MC = 2(21) + 2 = 42 + 2 = 44
(iii) x = 50,
MC. = 2(50) + 2 = 100+2 = 102
Posted by Tejasni Somasundaram 1 year, 11 months ago
- 1 answers
Preeti Dabral 1 year, 11 months ago
{tex}\begin{aligned} & \text { We know } \mathrm{V}=\frac{\mathrm{A}}{\mathrm{r}}\left[1-(1+\mathrm{r})^{-\mathrm{n}}\right] \\ & \text { Thus } 30000=\frac{\mathrm{A}}{0.12}\left[1-(1+0.12)^{-20}\right] \\ & \Rightarrow \mathrm{A}=\frac{30000 \times 0.12}{\left[1-(1+0.12)^{-20}\right]} \\ & \Rightarrow \mathrm{A}=\frac{3600}{\left[1-(1.12)^{-20}\right]} \\ & \Rightarrow \mathrm{A}=\text { Rs. } 4016.76 \end{aligned}{/tex}
Posted by Krishna Goswami Xii B 2 years, 8 months ago
- 2 answers
Tanushri Bhayani 1 year, 8 months ago
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- 0 answers
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- 1 answers
Posted by Sushma Mittal 1 year, 11 months ago
- 1 answers
Preeti Dabral 1 year, 11 months ago
74 - 120
= 48
62 ÷ 11
= 5
48 + 5
= 53
therefore, the test compared to the other volunteer
= 53
Posted by Khushi Thakur 3 years, 1 month ago
- 1 answers
Gembali Harsha Vardhan 3 years, 1 month ago
Posted by Deepansh Jayara 3 years, 1 month ago
- 1 answers
Gembali Harsha Vardhan 3 years, 1 month ago
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Preeti Dabral 1 year, 8 months ago
Answer is given below:
1Thank You