i need help with these algebra problems..i have no idea how to solve them?????!!!!???????

Question

Define a variable and write a quadratic function to describe each situation. Then find the two numbers that maximize or minimize each product
a- find the maximum product of two numbers whose sum is 80
b-find the minimum product of 2 numbers whose difference is 10


A store manager receives 100 dresses with a suggested retail price of $70. The projected sales are $20 dresses per week. Based on past experience, the store manager knows that sales will increase by 2 dresses per week for every $5 decrease in price.

a- find a function that gives the store's weekly revenue as a function of the number of $5 decreases
b-what price maximizes weekly revenue?

Answer 1

Define a variable and write a quadratic function to describe each situation. Then find the two numbers that maximize or minimize each product
a- find the maximum product of two numbers whose sum is 80

Let the numbers be a and b

so a + b = 80
So b = 80 - a

Product = a(80 - a)
= 80a - a²

= 1600 - 1600 + 80a - a² (making a perfect square)

= 1600 - (40 - a)²

So the maximum possible value of the product is 1600 and this occurs when a = 40 as (40 - a)² has a least value of 0 when a = 40

So the two numbers are 40 and 40

b-find the minimum product of 2 numbers whose difference is 10

Let the two numbers be a, b

Thus a - b = 10
So b = a - 10

Product = a(a - 10)
= a² - 10a
= a² - 10a + 25 - 25 (making a perfect square)
= (a - 5)² - 25

So the least value of the product is -25 and this occurs when a = 5 as (a - 5)² = 0, which is its least value when a = 5

So the two numbers are 5 and -5

A store manager receives 100 dresses with a suggested retail price of $70. The projected sales are 20 dresses per week. Based on past experience, the store manager knows that sales will increase by 2 dresses per week for every $5 decrease in price.

a- find a function that gives the store's weekly revenue as a function of the number of $5 decreases

Base revenue = 20 * $70
Let the number of $5 decreases be n
So price = $(70 - 5n)
This would result in the sale of (70 + 2n) dresses

So Revenue = (70 - 5n)(20 + 2n)

= 1400 - 100n + 140n - 10n²

= 1400 + 40n - 10n²

b-what price maximizes weekly revenue?

Total revenue = 1400 + 40n - 10n²

= 1400 - 10(n² - 4n)

= 1400 - 10(n² - 4n + 4 - 4) (making a perfect square)

= 1440 - 10(n - 2)²

So maximum revenue occurs when (n - 2)² = 0 ie when n = 2

So price that maximises revenue is when it is $(70 - 5*2) ie $60 per dress and the sales are (20 + 2*2) ie 24 per week. And also the maximum possible revenue from dress sales is $1440 per week

Answer 2

They are called optimization problems.

First:

a. find maximum product of 2 numbers whose sum is 80
1. Define variables

let x = one number
let y = other number

2. both numbers added together equals 80, so write an equation showing that.

80 = x + y

3. Solve the equation in terms of y (meaning, make y = something)

80 - x = y

there's the first value we need we need.

4. Now, find the equation for the optimization. We want the optimum product, lets call it P. That means:

P = x * y

5. substitute y for the value we got for it back in step 3, then multiply through

P = x (80 - x)
P = 80x - x²

So the quadratic equation needed is: P = -x² + 80x

6. We need to find the maximum, so we need to find out where the slope of the tangent line on this equation is equal to 0, or a horizontal line, because this indicates there's a peak.

so:

P' = -2x + 80

Set P' to 0

0 = -2x + 80

Solve.

-80 = -2x

x = 40

plug that value into original equation we got:

80 = x + y
80 = 40 + y

y = 40

So the two numbers are both 40.

these are long, so I can't do the rest for you right now, but post another question if you don't figure it out and I'll see what I can do.

Answer 3

A - 40 for both numbers - as for an equation x+y = 80
and something else
b -25 (+5 and -5) same reason - it's in the square

Answer 4

(x - 6)² = 0 (x - 6) (x - 6) = 0 set each and every aspect equivalent to 0 although the factors are a similar so: x - 6 = o x = 6 also 6 to the 2d ability is 36 no longer 12. 2d ability ability sq., no longer circumstances 2. on your equation you sq. the completed parentheses no longer the guy words.

Answer 5

do your own freakin homework. your not going to learn it this way