Answer:
a) x > − 24
b) x < − 3
c) q < 56
Step-by-step explanation:
a) −2/5x−9<9/10
<=> 2/5x + 9 > − 9/10
<=> 2/5x > − 9/10 − 9
<=> 2×2/2×5x > − 9/10 − 9×10/10
<=> 4/10x > − 99/10
<=> x > − 99/4
<=> x > − 24
b) 4x+6<−6
<=> 4x < − 6 − 6
<=> 4x < − 12
<=> x < − 12/4
<=> x < − 3
c) q+12−2(q−22)>0
<=> q+12−2q −2×(−22)>0
<=> (q−2q) + (12+ 44) >0
<=> −q + 56 >0
<=> q < 56
Answer:
In order to maximize the last equation we can derivate the function in term of x and we got:
And setting this derivate equal to 0 we got:
And solving for x we got:
And for this case the value that maximize the profit would be x =95 and the corresponding profit would be:
Step-by-step explanation:
For this case we have the following function for the profit:
And we can rewrite this expression like this:
In order to maximize the last equation we can derivate the function in term of x and we got:
And setting this derivate equal to 0 we got:
And solving for x we got:
And for this case the value that maximize the profit would be x =95 and the corresponding profit would be: