Respuesta :
Answer:
[tex]x_1 =2 , x_2=7[/tex]
Step-by-step explanation:
Consider the revenue function given by [tex]R(x_1,x_2) = -5x_1^2-8x_2^2 -2x_1x_2+34x_1+116x_2[/tex]. We want to find the values of each of the variables such that the gradient( i.e the first partial derivatives of the function) is 0. Then, we have the following (the explicit calculations of both derivatives are omitted).
[tex]\frac{dR}{dx_1} = -10x_1-2x_2+34 =0[/tex]
[tex]\frac{dR}{dx_2} = -16x_2-2x_1+116 =0[/tex]
From the first equation, we get, [tex]x_2 = \frac{-10x_1+34}{2}[/tex].If we replace that in the second equation, we get
[tex]-16\frac{-10x_1+34}{2} -2x_1+116=0= 80x_1-2x_1+116-272= 78x_1-156[/tex]
From where we get that [tex]x_1 = \frac{156}{78}=2[/tex]. If we replace that in the first equation, we get
[tex]x_2 = \frac{-10\cdot 2 +34}{2}=\frac{14}{2} = 7[/tex]
So, the critical point is [tex](x_1,x_2) = (2,7)[/tex]. We must check that it is a maximum. To do so, we will use the Hessian criteria. To do so, we must calculate the second derivatives and the crossed derivatives and check if the criteria is fulfilled in order for it to be a maximum. We get that
[tex] \frac{d^2R}{dx_1dx_2}= -2 = \frac{d^2R}{dx_2dx_1}[/tex]
[tex]\frac{d^2R}{dx_{1}^2}=-10, \frac{d^2R}{dx_{2}^2}=-16[/tex]
We have the following matrix,
[tex] \left[\begin{matrix} -10 & -2 \\ -2 & -16\end{matrix}\right][/tex].
Recall that the Hessian criteria says that, for the point to be a maximum, the determinant of the whole matrix should be positive and the element of the matrix that is in the upper left corner should be negative. Note that the determinant of the matrix is [tex](-10)\cdot (-16) - (-2)(-2) = 156>0[/tex] and that -10<0. Hence, the criteria is fulfilled and the critical point is a maximum
Answer:
x1 = 2000units, x2 = 7000units
Step-by-step explanation:
Given the total revenue in dollars of a company from x1 units of running shoes and x2 units of basketball shoes as;
R(x1,x2) = -5x1² − 8x2² − 2x1x2 + 34x1 + 116x2
The revenue function for running shoes will be gotten by differentiating the total revenue function with respect to x1 to have;
Rx1 = -10x1 - 2x2 + 34
Similarly to get the revenue function for units of basketball. We will have;
Rx2 = -16x2 - 2x1 + 116
Both products generates maximum/minimum revenue at their turning point i.e when Rx1 = 0 and Rx2 = 0
If Rx1 = 0 and Rx2 = 0 then;
Rx1 = -10x1 - 2x2 + 34 = 0... (1)
Rx2 = -16x2 - 2x1 + 116 ... (2)
Solving equation 1 and 2 simultaneously to get x1 and x2 at maximum revenue;
Using elimination method;
-10x1 - 2x2 = -34... (1) × 1
-16x2 - 2x1 = -116.. (2) × 5
The equations becomes;
-10x1 - 2x2 = -34
-80x2-10x1 = -580
Subtracting the resulting simultaneous equations will give;
-2x2+80x2 = -34+580
78x2 = 546
x2 = 546/78
x2 = 7
Substituting x2 = 7 into any of the equation we will have;
-10x1 - 2(7) = -34
-10x1 - 14= -34
-10x1 = -34+14
-10x1 = -20
x1 = 2
(x1,x2) = (2,7) will be our critical point
We must check if the critical point is at maximum first before we can find the maximum revenue generated for each product
Condition for the point to be maximum;
Rx1x1 < 0 and Rx1x1Rx2x2 - (Rx1x2)² >0
Rx1x1 = -10, Rx2x2 = -16, Rx1x2 = -2
It can been seen from the values that
Rx1x1 = -10<0
Also Rx1x1Rx2x2 - (Rx1x2) = -10(-16)-(-2)²
= 160-4
= 156>0
Since both conditions are met then the critical point is a maximum point. We can the calculate our maximum revenue for both products
Since x1 and x2 are in thousands of units,
x1 = 2×1000
x1 = 2000units
x2 = 7×1000
x2 = 7000units
The values of x1 and x2 needed so as to maximize the revenue is x1 = 2000units, x2 = 7000units
