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If the objective is to maximize the power dissipated on the load \$R\$. Use,

\$P_R = \frac{U^2 \cdot R}{(R + 3)^2 + 4^2} = \frac{U^2 \cdot R}{ R^2+6R+9+16} = \frac{U^2}{ R+6+25/R}\$

Then, derivedifferentiate the expression to find its maximum, that is, when \$R+6+25/R\$ reaches its minimum.

If you are actually looking for maximizing the power dissipated on both resistors, \$R\$ and \$ 3 \Omega \$, use,

\$P_{total} = \frac{U^2 \cdot (R+3)}{(R + 3)^2 + 4^2} = \frac{U^2}{\frac{(R + 3)^2 + 16}{R+3}} = \frac{U^2}{R + 3 + \frac{16}{R+3}}\$

Then, find when \$R + 3 + \frac{16}{R+3}\$ is at its smallest value.

If the objective is to maximize the power dissipated on the load \$R\$. Use,

\$P_R = \frac{U^2 \cdot R}{(R + 3)^2 + 4^2} = \frac{U^2 \cdot R}{ R^2+6R+9+16} = \frac{U^2}{ R+6+25/R}\$

Then, derive the expression to find its maximum, that is, when \$R+6+25/R\$ reaches its minimum.

If you are actually looking for maximizing the power dissipated on both resistors, \$R\$ and \$ 3 \Omega \$, use,

\$P_{total} = \frac{U^2 \cdot (R+3)}{(R + 3)^2 + 4^2} = \frac{U^2}{\frac{(R + 3)^2 + 16}{R+3}} = \frac{U^2}{R + 3 + \frac{16}{R+3}}\$

Then, find when \$R + 3 + \frac{16}{R+3}\$ is at its smallest value.

If the objective is to maximize the power dissipated on the load \$R\$. Use,

\$P_R = \frac{U^2 \cdot R}{(R + 3)^2 + 4^2} = \frac{U^2 \cdot R}{ R^2+6R+9+16} = \frac{U^2}{ R+6+25/R}\$

Then, differentiate the expression to find its maximum, that is, when \$R+6+25/R\$ reaches its minimum.

If you are actually looking for maximizing the power dissipated on both resistors, \$R\$ and \$ 3 \Omega \$, use,

\$P_{total} = \frac{U^2 \cdot (R+3)}{(R + 3)^2 + 4^2} = \frac{U^2}{\frac{(R + 3)^2 + 16}{R+3}} = \frac{U^2}{R + 3 + \frac{16}{R+3}}\$

Then, find when \$R + 3 + \frac{16}{R+3}\$ is at its smallest value.

Added the derivation for the second expression
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jDAQ
  • 2.6k
  • 1
  • 10
  • 19

If the objective is to maximize the power dissipated on the load \$R\$. Use,

\$P_R = \frac{U^2 \cdot R}{(R + 3)^2 + 4^2} = \frac{U^2 \cdot R}{ R^2+6R+9+16} = \frac{U^2}{ R+6+25/R}\$ Then

Then, derive the expression to find its maximum, that is, when \$R+6+25/R\$ reaches its minimum.

If you are actually looking for maximizing the power dissipated on both resistors, \$R\$ and \$ 3 \Omega \$, use,

\$P_R = \frac{U^2 \cdot (R+3)}{(R + 3)^2 + 4^2}\$\$P_{total} = \frac{U^2 \cdot (R+3)}{(R + 3)^2 + 4^2} = \frac{U^2}{\frac{(R + 3)^2 + 16}{R+3}} = \frac{U^2}{R + 3 + \frac{16}{R+3}}\$

Then, find when \$R + 3 + \frac{16}{R+3}\$ is at its smallest value.

If the objective is to maximize the power dissipated on the load \$R\$. Use,

\$P_R = \frac{U^2 \cdot R}{(R + 3)^2 + 4^2} = \frac{U^2 \cdot R}{ R^2+6R+9+16} = \frac{U^2}{ R+6+25/R}\$ Then, derive the expression to find its maximum, that is, when \$R+6+25/R\$ reaches its minimum.

If you are actually looking for maximizing the power dissipated on both resistors, \$R\$ and \$ 3 \Omega \$, use,

\$P_R = \frac{U^2 \cdot (R+3)}{(R + 3)^2 + 4^2}\$

If the objective is to maximize the power dissipated on the load \$R\$. Use,

\$P_R = \frac{U^2 \cdot R}{(R + 3)^2 + 4^2} = \frac{U^2 \cdot R}{ R^2+6R+9+16} = \frac{U^2}{ R+6+25/R}\$

Then, derive the expression to find its maximum, that is, when \$R+6+25/R\$ reaches its minimum.

If you are actually looking for maximizing the power dissipated on both resistors, \$R\$ and \$ 3 \Omega \$, use,

\$P_{total} = \frac{U^2 \cdot (R+3)}{(R + 3)^2 + 4^2} = \frac{U^2}{\frac{(R + 3)^2 + 16}{R+3}} = \frac{U^2}{R + 3 + \frac{16}{R+3}}\$

Then, find when \$R + 3 + \frac{16}{R+3}\$ is at its smallest value.

Source Link
jDAQ
  • 2.6k
  • 1
  • 10
  • 19

If the objective is to maximize the power dissipated on the load \$R\$. Use,

\$P_R = \frac{U^2 \cdot R}{(R + 3)^2 + 4^2} = \frac{U^2 \cdot R}{ R^2+6R+9+16} = \frac{U^2}{ R+6+25/R}\$ Then, derive the expression to find its maximum, that is, when \$R+6+25/R\$ reaches its minimum.

If you are actually looking for maximizing the power dissipated on both resistors, \$R\$ and \$ 3 \Omega \$, use,

\$P_R = \frac{U^2 \cdot (R+3)}{(R + 3)^2 + 4^2}\$