Find. Which bit (A or B) gives the lower cost per foot drilled, run to the end of its rated life.
Approach. Integrate the exponential ROP law over each bit's rated life to get the footage drilled and the out-depth $D_{out}$, add trip and connection time to the drilling time for a total run cost, and compare cost/ft.
Integrate the ROP equation. Separating variables, $e^{0.00025D}dD=300\,dt$, and integrating from $D_{in}$ to $D_{out}$ over the bit's life $t_d$: $$e^{0.00025D_{out}}=e^{0.00025D_{in}}+0.075\,t_d$$ with $e^{0.00025(12{,}000)}=e^{3}=20.086$.
Bit A ($t_d=40$ hr). $e^{0.00025D_{out}}=20.086+0.075(40)=23.086 \Rightarrow D_{out}=\ln(23.086)/0.00025$, so $\boxed{D_{out,A}=12{,}556.8\ \text{ft}}$, footage $\Delta D_A=556.8$ ft.
Bit B ($t_d=25$ hr). $e^{0.00025D_{out}}=20.086+0.075(25)=21.961 \Rightarrow D_{out}=\ln(21.961)/0.00025$, so $\boxed{D_{out,B}=12{,}357.0\ \text{ft}}$, footage $\Delta D_B=357.0$ ft.
Trip and connection time, each bit. $t_t=0.001D_{out}$; $t_c=(\Delta D/30)(2/60)$ hr. Bit A: $t_{t,A}=12.557$ hr, $t_{c,A}=0.619$ hr, total run time $=40+12.557+0.619=53.18$ hr. Bit B: $t_{t,B}=12.357$ hr, $t_{c,B}=0.397$ hr, total run time $=25+12.357+0.397=37.75$ hr.
Cost per foot. $C/ft=\dfrac{C_{bit}+C_R(t_d+t_t+t_c)}{\Delta D}$. Bit A: $C/ft=\dfrac{25{,}000+1{,}000(53.18)}{556.8}$, so $\boxed{(C/ft)_A=\$140.40/\text{ft}}$. Bit B: $C/ft=\dfrac{8{,}000+1{,}000(37.75)}{357.0}$, so $\boxed{(C/ft)_B=\$128.17/\text{ft}}$.
Decision. $128.17<140.40$, so $\boxed{\text{select Bit B}}$ — despite its shorter life and lower up-front cost, it drills at a lower cost per foot than the longer-lived, more expensive Bit A.
Check: each bit is assumed run to the full end of its rated life before tripping (the standard "minimum cost per foot" bit-economics convention when no premature-pull condition is stated).
