Respuesta :
Answer:
Value of v that minimizes E is v = 3u/2
Step-by-step explanation:
We are given that;
E(v) = av³L/(v-u)
Now, using the quotient rule, we have;
dE/dv = [(v-u)•3av²L - av³L(1)]/(v - u)²
Expanding and equating to zero, we have;
[3av³L - 3av²uL - av³L]/(v - u)² = 0
This gives;
(2av³L - 3av²uL)/(v-u)² = 0
Multiply both sides by (v-u)² to give;
(2av³L - 3av²uL) = 0
Thus, 2av³L = 3av²uL
Like terms cancel to give;
2v = 3u
Thus, v = 3u/2
To minimize a function, means that we want to get the minimum value of the function.
The value of v, that minimizes E is: [tex]\mathbf{v = \frac{3u}{2}}[/tex]
The function is given as:
[tex]\mathbf{E(v) = \frac{av^3L}{v - u}}[/tex]
First, we differentiate E(v) with respect to E using the quotient rule.
Quotient rule states that:
[tex]\mathbf{y' = \frac{V u' - Uv'}{v^2}}[/tex]
Using the above rule, we have:
[tex]\mathbf{E' = \frac{(v - u) \times 3av^2L - av^3L \times (-u)}{(v - u)^2}}[/tex]
To minimize the function, we set the equation to 0
[tex]\mathbf{\frac{(v - u) \times 3av^2L - av^3L \times (1)}{(v - u)^2} = 0}[/tex]
Cross multiply
[tex]\mathbf{(v - u) \times 3av^2L - av^3L = 0}[/tex]
Rewrite as:
[tex]\mathbf{(v - u) \times 3av^2L = av^3L }[/tex]
Divide through by av^2L
[tex]\mathbf{(v - u) \times 3 = v }[/tex]
Open brackets
[tex]\mathbf{3v - 3u = v}[/tex]
Collect like terms
[tex]\mathbf{3v - v = 3u}[/tex]
[tex]\mathbf{2v = 3u}[/tex]
Divide both sides by 2
[tex]\mathbf{v = \frac{3u}{2}}[/tex]
Hence, the value of v, that minimizes E is: [tex]\mathbf{v = \frac{3u}{2}}[/tex]
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