3-SAT ≤p TSP
If P ¹ NP, then no NP-complete problem can be solved in polynomial time.
both the statements are true.
<u>Explanation:</u>
- 3-SAT ≤p TSP due to any complete problem of NP to other problem by exits of reductions.
- If P ¹ NP, then 3-SAT ≤p 2-SAT are the polynomial time algorithm are not for 3-SAT. In P, 2-SAT is found, 3- SAT polynomial time algorithm implies the exit of reductions. 3 SAT does not have polynomial time algorithm when P≠NP.
- If P ¹ NP, then no NP-complete problem can be solved in polynomial time. because for the NP complete problem individually gets the polynomial time algorithm for the others. It may be in P for all the problems, the implication of latter is P≠NP.
Answer:
15.64 MW
Explanation:
The computation of value of X that gives maximum profit is shown below:-
Profit = Revenue - Cost
= 15x - 0.2x 2 - 12 - 0.3x - 0.27x 2
= 14.7x - .47x^2 - 12
After solving the above equation we will get maximum differentiate for profit that is
14.7 - 0.94x = 0
So,
x = 15.64 MW
Therefore for computing the value of X that gives maximum profit we simply solve the above equation.
This question is incomplete, the complete question is;
For a steel alloy it has been determined that a carburizing heat treatment of 11.3 h duration at Temperature T1 will raise the carbon concentration to 0.44 wt% at a point 1.8 mm from the surface. A separate experiment is performed at T2 that doubles the diffusion coefficient for carbon in steel.
Estimate the time necessary to achieve the same concentration at a 4.9 mm position for an identical steel and at the same carburizing temperature T2.
Answer:
the required time to achieve the same concentration at a 4.9 is 83.733 hrs
Explanation:
Given the data in the question;
treatment time t₁ = 11.3 hours
Carbon concentration = 0.444 wt%
thickness at surface x₁ = 1.8 mm = 0.0018 m
thickness at identical steel x₂ = 4.9 mm = 0.0049 m
Now, Using Fick's second law inform of diffusion
/ Dt = constant
where D is constant
then
/ t = constant
/ t₁ =
/ t₂
t₂ = t₁
t₂ = t₁
/ 
t₂ = (
/
)t₁
t₂ =
/
× t₁
so we substitute
t₂ =
0.0049 / 0.0018
× 11.3 hrs
t₂ = 7.41 × 11.3 hrs
t₂ = 83.733 hrs
Therefore, the required time to achieve the same concentration at a 4.9 is 83.733 hrs