How Should L and A Change to Adjust Power Dissipation and Current in a Wire?

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To adjust power dissipation and current in a wire, the relationship between resistance, resistivity, length, and cross-sectional area must be considered. The resistance is defined as R = ρL/A, and power is calculated using P = IV or P = V²/R. When the wire is stretched, its volume remains constant, meaning that as the cross-sectional area A decreases, the length L increases. To achieve a power increase of 35 times and a current increase of 3 times, both L and A must be recalibrated while maintaining the wire's volume. Understanding these relationships is crucial for determining the new dimensions of the wire.
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A potential difference V is applied to a wire of cross section A, length L, and resistivity p. You want to change the applied potential difference and stretch the wire so that the energy dissipation rate is multiplied by 35 and the current is multiplied by 3. What should be the new values of L and A [in relationship to the old values]?

I know that R = pL/A and P = IV (etc.), but it doesn't make sense that they'd want specific answers for both L and A. Can't either one change without the other and still affect the power?
 
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HINT: The volume of the metal does not change upon stretching.
 
So then volume V = AL. And if you're stretching it then A gets smaller and L gets bigger. But what does volume have to do with power?
 
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Fine. It doesn't matter anymore. Thanks.
 
Resistance can be found from L, A, and rho. Power can be found with P = IV = V^2/R.

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