How do you solve the following Maximum error problem via Derivatives?
The volume of a certain gas (in liters)is related to pressure P (in atmospheres) by the formula PV= 18. Suppose that V = 4 with a possible error of +or-.5L
a) Find the Maximum error in P.
b) Find the Relative error in computing P.
Update:Ok now how do u solve b?
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Use the approximation ΔP = (dP/dv) ΔV
P = 18/V
dP/dV = -18/(V²)
ΔP = -18/(V²) * ΔV = ± 18*0.5/16 = ± 9/16 atm (~0.6 atm)
x+y=18 y=18-x Maximize x(18-x)^2 permit p = x(18-x)^2 = x(324-36x+x^2) = 324x-36x^2+x^3 dp/dx = 324-72x+3x^2 = 0 equate the 1st spinoff to 0 and remedy for x 3x^2-72x+324=0 x^2-24+108=0 (x-18)(x-6)=0 x=18 or x=6 the 1st quantity is 6 and the 2d quantity is eighteen-(6)=12 d^2p/dx^2 = 6x-seventy two If x=6, 6x-seventy two < 0, verifies the product is optimal. the optimal product is (6)(12))^2 = 864