Find The Exact Location Of All Relative And Absolute Ext

Find The Exact Location Of All The Relative And Absolute Extrema Of Th

The problem requires identifying the relative and absolute extrema of the function g(x) = 6x³ − 72x within the domain [−4, 4]. Furthermore, the task involves classifying each extremum (whether it is a relative minimum, maximum, absolute minimum, or maximum) and providing their precise coordinates (x, y). Additional questions involve optimization for fencing costs in a rectangular garden, analyzing the acceleration of a point’s position over time, calculating the rate of change of a balloon's radius at the moment it pops, and examining the price elasticity of demand for a product. Each problem requires mathematical analysis, including derivatives, optimization, and application of related formulas, to find exact solutions and interpret their economic or physical significance.

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The first part of the assignment involves finding the critical and extremal points of the cubic function g(x) = 6x³ − 72x over the closed interval [−4, 4]. To do this, we start by calculating the first derivative of g(x) to locate critical points where the derivative equals zero or is undefined:

g'(x) = d/dx [6x³ − 72x] = 18x² − 72. Setting g'(x) = 0 gives:

18x² − 72 = 0, which simplifies to x² = 4, leading to x = ±2. These critical points are potential candidates for relative extrema. Next, we evaluate g at these points and at the endpoints of the domain to find the absolute extrema:

g(−4) = 6(−4)³ − 72(−4) = 6(−64) + 288 = −384 + 288 = −96.

g(−2) = 6(−2)³ − 72(−2) = 6(−8) + 144 = −48 + 144 = 96.

g(2) = 6(2)³ − 72(2) = 6(8) − 144 = 48 − 144 = -96.

g(4) = 6(4)³ − 72(4) = 6(64) − 288 = 384 − 288 = 96.

Analysis of the critical points' nature using the second derivative:

g''(x) = d/dx [18x² − 72] = 36x. At x = −2, g''(−2) = −72 0, indicating a local minimum at (2, -96). The function attains its absolute maximum value of 96 at (−2, 96) and (4, 96), and its absolute minimum value of -96 at (2, -96). Since the domain is closed and the function is continuous, these are also the global extrema.

The extremal points are:

  • At (−2, 96): Relative and absolute maximum.
  • At (2, -96): Relative and absolute minimum.
  • At (−4, −96): Endpoint, not an extremum, but value considered for global maximum/minimum.
  • At (4, 96): Endpoint, again extremum candidate; actual maximum due to earlier calculation.

The second part of the problem involves minimizing fencing costs for a rectangular garden with an area of 169 sq. ft., attached to the house forming the northern boundary. Let the length of the side perpendicular to the house be x, and the side parallel to the house be y. Because the house forms the northern boundary, fencing is needed only for the southern, eastern, and western sides, with costs per foot specified:

Area constraint: xy = 169 → y = 169/x.

Cost function: C(x) = 4(y) + 2(x + y) = 4(169/x) + 2(x + 169/x) = (4 * 169)/x + 2x + 2(169/x).

Simplifying, C(x) = 676/x + 2x + 338/x = (676 + 338)/x + 2x = 1014/x + 2x.

To minimize C(x), differentiate with respect to x:

C'(x) = -1014/x² + 2.

Set C'(x) = 0 to find critical points:

-1014/x² + 2 = 0 → 2 = 1014/x² → x² = 1014/2 = 507.

Thus, x = √507 ≈ 22.52 ft. Correspondingly, y = 169/x ≈ 169/22.52 ≈ 7.5 ft. The minimum fencing cost is obtained at these dimensions: south fence length is y ≈ 7.5 ft, and east and west fences each are x ≈ 22.52 ft.

The next part involves the analysis of a particle's position s(t) = 3/t − 3/t². First, the acceleration s''(t):

Compute first derivative s'(t):

s'(t) = d/dt [3/t − 3/t²] = -3/t² + 6/t³.

Second derivative s''(t):

s''(t) = d/dt [-3/t² + 6/t³] = 6/t³ − 18/t⁴.

At t=2, s''(2) = 6/8 − 18/16 = 0.75 − 1.125 = -0.375 ft/sec². This indicates the acceleration at t=2.

Finally, considering the spherical balloon whose volume V = (4/3)πr³, and the balloon pops when r = 6 cm. Given that dV/dt = 14 cm³/sec, the rate at which the radius r is growing at that moment involves differentiating volume with respect to time:

dV/dt = 4πr² dr/dt → dr/dt = dV/dt / (4πr²) = 14 / (4π * 36) ≈ 14 / (144π) ≈ 14 / 452.39 ≈ 0.0309 cm/sec.

Thus, at the moment when r=6, the radius is increasing at approximately 0.031 cm/sec.

The last part involves the demand function q = 1120 − 10p, where p is price per orange in dollars. To find the price elasticity of demand when p = $32, we calculate:

Elasticity E = (dq/dp) (p / q) = (−10) (32 / q).

At p = 32, q = 1120 − 10(32) = 1120 − 320 = 800. Therefore, E = (−10) (32 / 800) = -10 0.04 = -0.4. The elasticity magnitude is 0.4, indicating inelastic demand.

This means that a 1% increase in price will lead to approximately a 0.4% decrease in quantity demanded. To maximize revenue R = p q = p (1120 − 10p), differentiate R with respect to p:

dR/dp = 1120 − 20p. Set dR/dp = 0 to find the optimal price:

1120 − 20p = 0 → p = 1120/20 = $56.

Maximum revenue is then R = 56 (1120 − 1056) = 56 (1120 − 560) = 56 560 = $31,360.

In conclusion, every aspect of the problem involves applying calculus concepts such as derivatives for finding extrema and rates of change, optimization for the fencing problem, and elasticity calculations for demand analysis. These mathematical techniques provide precise solutions that can inform real-world decisions, from designing cost-effective gardens to understanding market behaviors and physical phenomena.

References

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