A one-product company finds that its profit, P, in millions of dollars, is given by the following equation where a is the amount spent on advertising, in millions of dollars, and p is the price charged per item of the product, in dollars.

Question

A one-product company finds that its profit, P, in millions of dollars, is given by the following equation where a is the amount spent on advertising, in millions of dollars, and p is the price charged per item of the product, in dollars.

P(a,p)=2ap+240p-25p^2-1/10a^2p-90

Find the maximum value of P and the values of a and p at which it is attained.

The maximum value of P is attained when a is $ million and p is $

The maximum value of P is $
million.

Answer 1

To find the maximum value of P and the values of a and p at which it is attained, we need to find the critical points. To do this, we'll take partial derivatives of the profit equation P(a, p) with respect to a and p, and set them equal to zero.

Partial derivative with respect to a:
∂P/∂a = 2p - (1/5)ap = 0

Partial derivative with respect to p:
∂P/∂p = 2a + 240 - 50p = 0

From the first equation, we can rearrange to solve for a:
2p - (1/5)ap = 0
2p = (1/5)ap
10p = ap
a = 10

Now substitute this value of a into the second equation:
2(10) + 240 - 50p = 0
20 + 240 - 50p = 0
-50p = -260
p = 5.2

Therefore, the maximum value of P is attained when a is $10 million and p is $5.2.

To find the maximum value of P, substitute the values of a and p into the profit equation:
P(a, p) = 2ap + 240p - 25p^2 - (1/10)a^2p - 90
P(10, 5.2) = 2(10)(5.2) + 240(5.2) - 25(5.2)^2 - (1/10)(10)^2(5.2) - 90
P(10, 5.2) = 104 + 1248 - 676 - 52 - 90
P(10, 5.2) = 534

Therefore, the maximum value of P is $534 million.

Answer 2

To find the maximum value of P, we need to find the critical points of the profit function. The critical points occur where the derivative of P with respect to both a and p is equal to zero.

To find the partial derivatives of P, we differentiate the equation with respect to each variable:

∂P/∂a = 2p - (1/5)ap
∂P/∂p = 2a + 240 - 50p

Now, set both derivatives equal to zero and solve for a and p:

2p - (1/5)ap = 0
2a + 240 - 50p = 0

From the first equation, we can see that either p = 0 or a = 10p. Substituting this value of a into the second equation:

2(10p) + 240 - 50p = 0
20p + 240 - 50p = 0
-30p + 240 = 0
30p = 240
p = 8

Now that we have the value of p, we can substitute it back into the first equation to find the corresponding value of a:

2(8) - (1/5)a(8) = 0
16 - (8/5)a = 0
16 = (8/5)a
a = (5/8)(16)
a = 10

So, when a = 10 million dollars and p = 8 dollars, the profit function P is maximized. To find the maximum value of P, substitute these values into the profit equation:

P(a,p) = 2ap + 240p - 25p^2 - (1/10)a^2p - 90
P(10, 8) = 2(10)(8) + 240(8) - 25(8)^2 - (1/10)(10)^2(8) - 90
P(10, 8) = 160 + 1920 - 25(64) - 64 - 90
P(10, 8) = 160 + 1920 - 1600 - 64 - 90
P(10, 8) = 326

Therefore, the maximum value of P is $326 million.

Answer 3

A​ one-product company finds that its​ profit, P, in millions of​ dollars, is given by the following equation where a is the amount spent on​ advertising, in millions of​ dollars, and p is the price charged per item of the​ product, in dollars.

To find the maximum value of P and the values of a and p at which it is attained, we need to find the critical points. To do this, we'll take partial derivatives of the profit equation P(a, p) with respect to a and p, and set them equal to zero.

To find the maximum value of P, we need to find the critical points of the profit function. The critical points occur where the derivative of P with respect to both a and p is equal to zero.

Set the partial derivatives to zero for extrema. Both must be zero.

A one-product company finds that its profit, P, in millions of dollars, is given by the following equation where a is the amount spent on advertising, in millions of dollars, and p is the price charged per item of the product, in dollars.