1) Find two real numbers whose sum is 30 and whose product is maximized.
2) Find two numbers whose difference is 50 and whose product is minimized.

1) Both 15.

2) 25 and -25.

Step-by-step explanation:

1) Let the 2 numbers be x and 30 - x.

The product  = x(30 - x)

f(x) = x(30 - x)

f(x) = 30x - x^2

Finding  the derivative:

f'(x) = 30 - 2x

Finding the maximum:

30 - 2x = 0

x = 15.

This gives a maximum f(x) because f"(x) = -2.

So the numbers are  15 and 30 - 15 = 15.

2).  If one number is x the other is y.

x - y = 50

y = x - 50

The product =

x(x - 50)

= x^2 - 50x

Finding the derivative:

2x - 50 = 0   for  a minimum value.

2x = 50

x = 25.

So the numbers are  25 and 25-50 = -25.

l.jnl.j/n.m,lm,

step-by-step explanation:

step-by-step explanation: so we look at the formula un = a + (n-1)d

un is the amount of seats

n is the nth term aka row number

a is the original sequence/row number

d is the difference between each term

so in this a=22 d=6 and n can remain as n because we are simply formjng a rule

lets put these in the equation

un=22+(n-1)(6)

this is your answer to part a

however

lets double check that this is correct, we'll try putting it into the second term

un=a+(n-1)d

u2=22+(2-1)(6)

u2=22+6

u2=28

as the question said the second row has 28 seats so now we know the formula we made is correct.

as for part b

un=a+(n-1)

its asking us to find a row with 100 seats.

remember un was the amount of seata so all we do is substitute this in.

un=100 a=22 d=6

un=a+(n-1)d

100=22+(n-1)(6)

so now we simplify to find n which is the row number.

100-22=22+(n-1)(6)-22

78=(n-1)(6)

78/6=(n-1)(6)/6

13=n-1

13+1=n-1+1

14=n

so now we know that in the 14th row there is 100 seats.

hope this and that you understood! its my first time using the app or anyone out on here : p its also me to revise so a bunch! ask more questions if you dont understand ☺