What is the maximum value of the objective function, P, with the given constraints? P=25x+45y 4x+y≤16 x+y≤10 x≥0 y≥0

the maximum is 450 and the minimum is 0.

Step-by-step explanation:

We know that:

P(x, y) = 25*x + 45*y

First, is easy to see that as x and y increase, also does the value of P(x, y)

So we just need to find the largest and smallest possible values of these variables. We also can notice that the variable y is being multiplicated by a larger coefficient than x, so we prioritize larger values of y when we can.

We know that:

4x+y ≤ 16

x + y ≤10

x≥0

y≥0

Let's start with the second inequality, let's solve this for y:

y ≤ 10 - x

and from the first one we get:

y ≤ 16 - 4*x

Just to show that maximizing x does not work, let's do it:

from the second one, knowing that the minimum value of y is y = 0

we have that:

0 ≤ 16 - 4*x

Here the maximum value that y can take is x = 1

0 ≤ 16 - 4*1 = 0

So we can have the combination y = 0 and x = 1, when we maximize x (here we can see that we should not maximize x)

in this case we get:

P(1, 0) = 25*1 + 47*0 = 25

let's write again our inequalities:

y ≤ 10 - x

y ≤ 16 - 4*x

If now we take the minimum value of x, x = 0, we get:

y ≤ 10

y ≤ 16

Because the first one is more restrictive, we know that the maximum value that y can take (when x = 0) is y = 10

in this case we get:

P(0, 10) = 25*0 + 45*10 = 450

As expected, here is the actual maximum for the given restrictions.

For the minimum, we just need to take the two lowest possible values of x and y, which are the two given by the equalities on:

x≥0

y≥0

The smallest values are:

x = 0

y = 0

Replacing that in the equation we get:

P(0, 0) = 25*0 + 47*0 = 0

So the maximum is 450 and the minimum is 0. (with the given restrictions)

you have to put the graph on here with the question

step-by-step explanation:

the answer is a.

135 and 80

step-by-step explanation: