Respuesta :
Answer:
93 items should be manufacture so that the profit is maximum.
Step-by-step explanation:
Given that,
Total revenue from x items is
R(x)= 31 in (5x+1)
Total cost to produce x items is
[tex]C(x)=\frac x3[/tex]
Profit = revenue- cost
P(x)=31 ln (5x+1) - [tex]\frac x3[/tex]
Differentiating with respect to x
[tex]P'(x)= 31 \frac{5}{(5x+1)}-\frac13[/tex]
Again differentiating with respect to x
[tex]P''(x)= -31 \frac{25}{(5x+1)^2}[/tex]
First, we set P'(x) =0
[tex]31 \frac{5}{(5x+1)}-\frac13=0[/tex]
[tex]\Rightarrow \frac{155}{(5x+1)}-\frac13=0[/tex]
[tex]\Rightarrow \frac{155\times 3-1.(5x+1)}{3(5x+1)}=0[/tex]
⇒465 - 5x-1 =0
⇒5x= 464
[tex]\Rightarrow x=\frac{464}{5}[/tex]
[tex]\Rightarrow x=92.8[/tex]
[tex]\Rightarrow x\approx 93[/tex]
Now [tex]P''(x)|_{x=93}= -31 \frac{25}{(5\times 93+1)^2}<0[/tex]
So, at x=93 the profit is maximum.
93 items should be manufacture so that the profit is maximum.
So, approximately 93 times should be manufactured to maximize the profit.
To understand the calculations, check below
Maximization Problem:
A standard maximization problem is a linear programming problem in which we seek to maximize an objective function [tex]P = c_1x_1 +..+ c_nx_n[/tex] .
Given that,
[tex]R(x) = 31 ln (5x + 1)\\C(x)=\frac{x}{3}[/tex]
Now, solving the above functions as,
[tex]R(x)-C(x) = 31 ln (5x + 1)-\frac{x}{3}[/tex]
For [tex]R(x)-C(x)[/tex] to be maximum,
Let [tex]P(x)=R(x)-C(x)[/tex]
As we know for maximum profit [tex]P'(x)=0[/tex] then,
[tex]\frac{d}{dx}[ 31 ln (5x + 1)-\frac{x}{3} ] =0\\\frac{31\times 5}{5x+2} -\frac{1}{3} =0\\5x+2=465\\x=92.6\\x\approx 93[/tex]
Learn more about the topic Maximization Problem:
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