You run a small furniture business. You sign a deal with a customer to deliver up to 700 chairs, the exact number to be determined by the customer later. The price will be $150 per chair up to 600 chairs, and above 600 the price will be reduced by $0.25 per chair (on the whole order) for every additional chair over 600 ordered. What is the largest revenue your company can make under this deal?
I understand that the smallest revenue would be $0 since they could sell 0 chairs but I don't understand how to find the largest revenue.
I did R= (150-0.25x) (600+x)
so 90000-150x+150x-0.25x^2
so then 90000-0.25x^2 and you set that equal to 0... but then you just get 0 out of that? and that can't be the maximum revenue.
To find the largest revenue your company can make under this deal, we need to determine the optimal number of chairs to sell that maximizes the revenue. Let's break down the problem further.
Let:
x = number of chairs ordered above 600 (additional chairs)
y = number of chairs ordered up to 600 (base chairs)
The revenue function can be defined as follows:
R = (150 * y) + (150 - 0.25 * x) * x
To maximize the revenue, we need to find the value of x that results in the maximum R.
Since we know that the customer can order up to 700 chairs, the total number of chairs ordered can be expressed as:
Total number of chairs = y + x
Therefore, we have:
y + x ≤ 700
y ≤ 700 - x
Let's rewrite the revenue function to only include y:
R = (150 * (700 - x)) + (150 - 0.25 * x) * x
Now, we can simplify the equation and express it as a quadratic function:
R = -0.25x^2 + 150x + 105000
To find the maximum revenue, we need to find the peak of the quadratic function.
The x-value for the peak can be determined using the formula:
x = -b / (2a)
In this case, a = -0.25 and b = 150. Plugging in the values, we get:
x = -(150) / (2 * -0.25)
x = -150 / -0.5
x = 300
To find the corresponding y-value, we can substitute x = 300 into the constraint equation:
y = 700 - x
y = 700 - 300
y = 400
Therefore, to maximize revenue, you should sell 400 chairs up to 600 and an additional 300 chairs above 600. The largest revenue your company can make is:
R = (150 * 400) + (150 - 0.25 * 300) * 300
R = 60,000 + 67,500
R = $127,500
Hence, the largest revenue your company can make under this deal is $127,500.
To find the largest revenue your company can make under this deal, you need to optimize the revenue equation with respect to the number of chairs ordered. Let's break down the problem into two scenarios: ordering up to 600 chairs, and ordering more than 600 chairs.
1. Ordering up to 600 chairs:
For the first 600 chairs, the price per chair is $150. So, the revenue for this portion can be represented as 150x, where x is the number of chairs ordered up to 600.
2. Ordering more than 600 chairs:
For each additional chair ordered beyond 600, the price per chair is reduced by $0.25. So, the revenue for this portion can be represented as (150 - 0.25(x - 600))(x - 600), where x is the total number of chairs ordered.
To find the largest revenue, you need to maximize the total revenue, which is the sum of the revenue from both scenarios.
Let's simplify the equation:
Revenue = 150x + (150 - 0.25(x - 600))(x - 600)
Expanding and rearranging the equation:
Revenue = 150x + (150x - 0.25x^2 + 90,000 - 90x - 90,000)
Revenue = 300x - 0.25x^2
To find the maximum revenue, you need to find the value of x that maximizes the equation Revenue = 300x - 0.25x^2.
To do this, you can use calculus by taking the derivative of Revenue with respect to x and setting it equal to zero.
Let's find the derivative:
dRevenue/dx = 300 - 0.5x
Set dRevenue/dx equal to zero:
300 - 0.5x = 0
0.5x = 300
x = 600
Based on this calculation, the maximum revenue is achieved when the customer orders 600 chairs.
Therefore, the largest revenue your company can make under this deal is:
Revenue = 300(600) - 0.25(600^2) = $90,000.
Your function is correct.
Maximum revenue is at x=0.
That means that an order for 600 chairs produces the maximum revenue.
This would not be the case if the discount were applied only to the excess chairs. Then the revenue would be
R(x) = 600*150 + x(150-.25x)
Then the maximum revenue would be at x=300. But the maximum order is 700 chairs, or x=100.
So, either way the problem seems flawed in its design.