Hello! Can u help me with this please. I'm having a hard time solving this one but I don't know if I make it right. Please help me.
Suppose that 10,000𝑥 dollars is the inventory carried in a store
employing 𝑦 clerks, 𝑃 dollars is the weekly profit of the store, and
𝑃 = 3000 + 240𝑦 + 20𝑦 ( 𝑥 − 2𝑦 ) − 10(𝑥 − 12)²
where 15 ≤ 𝑥 ≤ 25 and 5 ≤ 𝑦 ≤ 12 . At present the inventory is
$180,000 and there are 8 clerks.
(a) Find the instantaneous rate of change of 𝑃 per unit change in 𝑥 if
𝑦 remains fixed at 8.
(b) Use the result of part (a) to find the approximate change in the
weekly profit if the inventory changes from $180,000 to $200,000 and the number of clerks remains fixed at 8.
(c) Find the instantaneous rate of change of 𝑃 per unit change in 𝑦 if 𝑥 remains fixed at 18.
(d) Use the result of part (c) to find the approximate change in the
weekly profit if the number of clerks is increased from 8 to 10 and the inventory remains fixed at $180,000.
p = 3000 + 240y + 20y (x-2y) - 10(x-12)^2
(a) dp/dx = 20*8x - 20(x-12)
(b) ∆p ≈ dp/dx ∆x
so, when the inventory changes from 180K to 200K, x changes from 18 to 20
∆p ≈ (160*18 - 20(18-12)) * (20-18) = ____
(c) p = 3000 + 240y + 20y (x-2y) - 10(x-12)^2
dp/dy = 240 + 20(18-2y) + 20y(-2)
(d) do as in (c)
Thank u so much. It means a lot.
Sure, I can help you with that! But before we dive into the math, let me warn you that my answers might contain a bit of clowning around. So, bear with me!
(a) To find the instantaneous rate of change of 𝑃 per unit change in 𝑥, we need to find the derivative of 𝑃 with respect to 𝑥 when 𝑦 is fixed at 8. Let's do it step by step:
First, we have 𝑃 = 3000 + 240𝑦 + 20𝑦 ( 𝑥 − 2𝑦 ) − 10(𝑥 − 12)².
To find the derivative with respect to 𝑥, we differentiate each term separately:
For the first term, the derivative of a constant term (3000) is 0. So, it vanishes.
For the second term, the derivative of 240𝑦 (since 𝑦 is fixed at 8) with respect to 𝑥 is 0. So, that also disappears.
For the third term, we have 20𝑦 ( 𝑥 − 2𝑦 ). To differentiate it with respect to 𝑥, treat 𝑦 as a constant. So, the derivative of 20𝑦 ( 𝑥 − 2𝑦 ) is just 20𝑦.
For the last term, we have -10(𝑥 − 12)². The derivative of (𝑥 − 12)² with respect to 𝑥 is 2(𝑥 − 12). But since 𝑦 is fixed at 8, this becomes 2(𝑥 − 12) multiplied by 0, which is 0.
So, the derivative of 𝑃 with respect to 𝑥 is 20𝑦.
(b) Now, let's move on to finding the approximate change in the weekly profit if the inventory changes from $180,000 to $200,000 while the number of clerks remains fixed at 8. We can use the result from part (a), where we found that the derivative of 𝑃 with respect to 𝑥 is 20𝑦.
Since 𝑦 remains fixed at 8, we can substitute 𝑦 = 8 into the derivative: 20(8) = 160.
So, the approximate change in the weekly profit is 160 dollars per unit change in 𝑥. Since the inventory changes from $180,000 to $200,000, the change in 𝑥 is 20,000 dollars. Thus, the approximate change in the weekly profit is 160 dollars multiplied by 20,000, which is $3,200,000.
(c) To find the instantaneous rate of change of 𝑃 per unit change in 𝑦, with 𝑥 fixed at 18, we need to find the derivative of 𝑃 with respect to 𝑦. Let's do it!
Differentiate 𝑃 = 3000 + 240𝑦 + 20𝑦 ( 𝑥 − 2𝑦 ) − 10(𝑥 − 12)², this time with respect to 𝑦:
The derivative of 3000 with respect to 𝑦 is 0. So, it disappears.
The derivative of 240𝑦 is 240.
The derivative of 20𝑦 ( 𝑥 − 2𝑦 ) with respect to 𝑦 is 20(𝑥 − 2𝑦) - 20(2) = 20𝑥 - 40𝑦.
The derivative of -10(𝑥 − 12)² with respect to 𝑦 is zero, since 𝑥 is fixed at 18.
Combining all these terms, we get the derivative of 𝑃 with respect to 𝑦 as 240 + 20𝑥 - 40𝑦.
(d) Finally, let's use the result from part (c) to find the approximate change in the weekly profit if the number of clerks is increased from 8 to 10 while the inventory remains fixed at $180,000.
We found that the derivative of 𝑃 with respect to 𝑦 is 240 + 20𝑥 - 40𝑦. Since 𝑥 remains fixed at 18, we can substitute 𝑥 = 18 into the derivative: 240 + 20(18) - 40(8) = 240 + 360 - 320 = 280.
So, the instantaneous rate of change of 𝑃 per unit change in 𝑦 is 280 dollars.
Since the number of clerks increases from 8 to 10, the change in 𝑦 is 2. Thus, the approximate change in the weekly profit is 280 dollars multiplied by 2, which is $560.
I hope that helps! If you have any more questions or if you'd like me to add a little more humor to my answers, feel free to ask!
Yes, I can help you with this problem. Let's break it down step by step.
(a) To find the instantaneous rate of change of 𝑃 per unit change in 𝑥 when 𝑦 remains fixed at 8, we need to find the derivative of 𝑃 with respect to 𝑥.
To do this, we differentiate the given expression for 𝑃 with respect to 𝑥, treating 𝑦 as a constant:
𝑃 = 3000 + 240𝑦 + 20𝑦 ( 𝑥 - 2𝑦 ) - 10(𝑥 - 12)²
Differentiating 𝑃 with respect to 𝑥, we get:
𝑑𝑃/𝑑𝑥 = 20𝑦 - 20(𝑥 - 12) * 2
Simplifying further, we have:
𝑑𝑃/𝑑𝑥 = 20𝑦 - 40(𝑥 - 12)
Now, substitute 𝑦 = 8 into the equation:
𝑑𝑃/𝑑𝑥 = 20(8) - 40(𝑥 - 12)
Simplifying this, we get:
𝑑𝑃/𝑑𝑥 = 160 - 40𝑥 + 480
𝑑𝑃/𝑑𝑥 = -40𝑥 + 640
So, the instantaneous rate of change of 𝑃 per unit change in 𝑥, when 𝑦 remains fixed at 8, is given by -40𝑥 + 640.
(b) To find the approximate change in the weekly profit if the inventory changes from $180,000 to $200,000 and the number of clerks remains fixed at 8, we can use the result from part (a).
The change in 𝑃 is given by 𝛥𝑃 = 𝑑𝑃/𝑑𝑥 * 𝛥𝑥, where 𝛥𝑥 is the change in 𝑥.
Here, 𝛥𝑥 = $200,000 - $180,000 = $20,000.
Substituting 𝑦 = 8 into the equation from part (a), we have:
𝑑𝑃/𝑑𝑥 = -40𝑥 + 640
𝛥𝑃 = (-40𝑥 + 640) * 𝛥𝑥
𝛥𝑃 = (-40𝑥 + 640) * $20,000
Now, substitute 𝑥 = 180,000 into the equation:
𝛥𝑃 = (-40(180,000) + 640) * $20,000
Simplifying this expression will give us the approximate change in the weekly profit 𝑃 when the inventory changes from $180,000 to $200,000 and the number of clerks remains fixed at 8.
(c) To find the instantaneous rate of change of 𝑃 per unit change in 𝑦 when 𝑥 remains fixed at 18, we again need to find the derivative of 𝑃 with respect to 𝑦.
Using the same expression for 𝑃, differentiate 𝑃 with respect to 𝑦, treating 𝑥 as a constant:
𝑑𝑃/𝑑𝑦 = 240 + 20(𝑥 - 2𝑦)
Substitute 𝑥 = 18 into the equation:
𝑑𝑃/𝑑𝑦 = 240 + 20(18 - 2𝑦)
Simplifying this equation gives us the instantaneous rate of change of 𝑃 per unit change in 𝑦 when 𝑥 remains fixed at 18.
(d) To find the approximate change in the weekly profit if the number of clerks is increased from 8 to 10 and the inventory remains fixed at $180,000, we can use the result from part (c).
The change in 𝑃 is given by 𝛥𝑃 = 𝑑𝑃/𝑑𝑦 * 𝛥𝑦, where 𝛥𝑦 is the change in 𝑦.
Here, 𝛥𝑦 = 10 - 8 = 2.
Substituting 𝑥 = 18 into the equation from part (c), we have:
𝑑𝑃/𝑑𝑦 = 240 + 20(18 - 2𝑦)
𝛥𝑃 = (240 + 20(18 - 2𝑦)) * 𝛥𝑦
𝛥𝑃 = (240 + 20(18 - 2(8))) * 2
Simplifying this expression will give us the approximate change in the weekly profit 𝑃 when the number of clerks is increased from 8 to 10 and the inventory remains fixed at $180,000.
Sure, I can help you with that! Let's go through each part step by step.
(a) To find the instantaneous rate of change of 𝑃 per unit change in 𝑥 if 𝑦 remains fixed at 8, we need to find the derivative of 𝑃 with respect to 𝑥, and then evaluate it at 𝑦=8.
The given expression for 𝑃 is: 𝑃 = 3000 + 240𝑦 + 20𝑦 ( 𝑥 − 2𝑦 ) − 10(𝑥 − 12)²
To find the derivative, we can use the rules of differentiation. Taking the derivative of each term:
𝑑𝑃/𝑑𝑥 = 0 + 0 + 20𝑦 - 20(𝑥 - 12) * 2(𝑥 - 12)'
Simplifying, 𝑑𝑃/𝑑𝑥 = 20𝑦 - 40(𝑥 - 12)(𝑥 - 12)'
Now, substituting 𝑦=8 into this expression will give us the instantaneous rate of change of 𝑃 per unit change in 𝑥 when 𝑦 is fixed at 8. To find the value of 𝑥', we need to plug in the values given for 𝑥 and 𝑦 into the differentiated expression.
(b) To find the approximate change in the weekly profit if the inventory changes from $180,000 to $200,000 and the number of clerks remains fixed at 8, we can use the result obtained in part (a). We will multiply the value of 𝑥' from part (a) by the change in 𝑥, which is 𝛥𝑥 = 200,000 - 180,000.
Approximate change in weekly profit = 𝑑𝑃/𝑑𝑥 * 𝛥𝑥
(c) To find the instantaneous rate of change of 𝑃 per unit change in 𝑦 if 𝑥 remains fixed at 18, we need to find the derivative of 𝑃 with respect to 𝑦 and evaluate it at 𝑥=18.
Taking the derivative of 𝑃 with respect to 𝑦, we get:
𝑑𝑃/𝑑𝑦 = 0 + 240 + 20(18 - 2𝑦)
Simplifying, 𝑑𝑃/𝑑𝑦 = 240 + 20(18 - 2𝑦)
Now, substituting 𝑥=18, we can evaluate this expression.
(d) To find the approximate change in the weekly profit if the number of clerks is increased from 8 to 10 and the inventory remains fixed at $180,000, we can use the result obtained in part (c). We will multiply the value of 𝑦' from part (c) by the change in 𝑦, which is 𝛥𝑦 = 10 - 8.
Approximate change in weekly profit = 𝑑𝑃/𝑑𝑦 * 𝛥𝑦
These steps should help you solve the given problem. If you need any further clarification or assistance, please let me know!