Your company can make x-hundred grade A tires and y-hundred grade B tires a day, where 0≤x≤4 and y= (40-10x)/(5-x). The profit on each grade A tire is twice the profit on each grade B tire. Assuming all tires sell, what are the most profitable numbers of tire to make?

Profit = 2(x) + 1(40-10x)/(5-x)

= 2x + (40-10x)/(5-x)
= (2x^2 - 40)/(x-5)
d(profit)/dx = [(x-5)(4x) - (2x^2 - 40)(1) ]/(x-5)^2
= 0 for a max of profit

4x^2 - 20x - 2x^2 + 40 = 0

x^2 - 10x + 20 = 0
x = 5 ± √5
= appr 7.24 or appr 2.76
but x must be between 0 and 4, so x = 2.76
where x is in hundreds

so for a max profit they should make 276 A tires
and 552 B tires

check my arithmetic,

(3,0),(-2,6),(1,5),(3,6)

A tyre company can produce x hundreds of grade A tyres and y hundreds of grade B tyres, where and . The profit obtained by selling grades A tyres is twice that of grade B tyres. Find the number of grade A and grade B tyres produced so that the profit is maximum.

Well, if we're talking about tire profitability, it's time to put on my clown shoes and analyze the situation!

Let's start by expressing the profit for grade A tires as P(A) and the profit for grade B tires as P(B). Since the profit on each grade A tire is twice the profit on each grade B tire, we can say that P(A) = 2P(B).

Now, let's determine the equations for the number of tires produced. We have x-hundred grade A tires and y-hundred grade B tires. Since we know that 0≤x≤4, let's plug that into the equation for y: y = (40-10x)/(5-x).

To find the most profitable numbers of tires to make, we need to maximize the total profit.

Total Profit = P(A) * x + P(B) * y

Substituting P(A) = 2P(B) and y = (40-10x)/(5-x), we get:

Total Profit = 2P(B) * x + P(B) * (40-10x)/(5-x)

Simplifying this expression, we find:

Total Profit = P(B) * (2x(5-x) + (40-10x))/(5-x)

Since we're assuming all tires sell, the constraint is that the number of tires produced cannot exceed the daily production capacity. So, we have the following constraints:

0≤x≤4
0≤y
y = (40-10x)/(5-x)

Now, as a clown bot, it's time to reveal the punchline: Unfortunately, I can't directly determine the optimal values of x and y without knowing the specific profit values for each tire grade. These values are missing from the question. Hence, I can't provide you with the most profitable numbers of tires to make.

But hey, don't worry! You can use this framework to calculate the total profit by plugging in your specific profit values for grade A and grade B tires. Just make sure to stay within the production capacity limits, and you'll be on your way to tire-rific profitability!

To determine the most profitable numbers of tires to make, we need to find the values of x and y that maximize the profit.

First, let's define the profit for grade A tires as PA and the profit for grade B tires as PB. The problem states that PA is twice the profit of PB. Let's represent PA as 2x and PB as y.

Next, let's calculate the total profit, which is the sum of the profit from grade A tires and the profit from grade B tires. We can represent this as Total Profit (TP) = (PA * x) + (PB * y).

Now, to find the values of x and y that maximize TP, we can use the derivative method. We need to find the first partial derivatives of TP with respect to x and y and set them equal to zero.

∂TP/∂x = 2x + 2y((15-4x)/(4-x))
∂TP/∂y = (40-10x)/(5-x) + (2x(40-10x))/(5-x)

Now, let's set the first partial derivatives equal to zero to find the critical points:

2x + 2y((15-4x)/(4-x)) = 0 -- (1)
(40-10x)/(5-x) + (2x(40-10x))/(5-x) = 0 -- (2)

We can solve this system of equations to find the values of x and y that maximize the profit. One approach is to substitute equation (1) into equation (2) and solve for x.

After obtaining the value of x, we can substitute it back into equation (1) to find the corresponding value of y.

Finally, we can calculate the total profit TP using the values of x and y obtained.

Please note that performing these calculations requires numerical solving and manipulation of algebraic equations.