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Zorluk: ZorStationary Points, Maxima, and Minima

An open rectangular box with a square base of side length x cmx\text{ cm} is to be constructed such that its total surface area is 108 cm2108\text{ cm}^2. What is the maximum volume of the box in cm3\text{cm}^3?

  1. A
    54 cm354\text{ cm}^3
  2. B
    72 cm372\text{ cm}^3
  3. 108 cm3108\text{ cm}^3Cevap
  4. D
    432 cm3432\text{ cm}^3

Cevap

108 cm3108\text{ cm}^3
For an open box with a square base of side xx and height hh, total surface area is A=x2+4xh=108 cm2A = x^2 + 4xh = 108\text{ cm}^2. Solving for hh gives h=108−x24xh = \frac{108 - x^2}{4x}. Substituting hh into the volume formula gives V(x)=x2h=27x−14x3V(x) = x^2 h = 27x - \frac{1}{4}x^3. Differentiating gives dVdx=27−34x2\frac{dV}{dx} = 27 - \frac{3}{4}x^2. Setting dVdx=0\frac{dV}{dx} = 0 yields x=6x = 6. The second derivative d2Vdx2=−32x\frac{d^2V}{dx^2} = -\frac{3}{2}x evaluated at x=6x = 6 is −9<0-9 < 0, confirming a local maximum. Evaluating V(6)=27(6)−14(63)=162−54=108 cm3V(6) = 27(6) - \frac{1}{4}(6^3) = 162 - 54 = 108\text{ cm}^3 yields the correct maximum volume.

Adım Adım Çözüm

1
Set up the surface area equation for the open box and express height hh in terms of xx.
Surface area A=x2+4xh=108  ⟹  h=108−x24xA = x^2 + 4xh = 108 \implies h = \frac{108 - x^2}{4x}.
An open box with a square base has 1 base face (x2x^2) and 4 vertical side faces (xhxh).
2
Formulate the volume function V(x)V(x) in terms of xx.
V(x)=x2h=x2(108−x24x)=14(108x−x3)=27x−14x3V(x) = x^2 h = x^2 \left(\frac{108 - x^2}{4x}\right) = \frac{1}{4}(108x - x^3) = 27x - \frac{1}{4}x^3.
Substitute the expression for hh into the volume formula V=x2hV = x^2 h.
3
Find the critical point by differentiating V(x)V(x) with respect to xx and setting dVdx=0\frac{dV}{dx} = 0.
dVdx=27−34x2=0  ⟹  34x2=27  ⟹  x2=36  ⟹  x=6 cm\frac{dV}{dx} = 27 - \frac{3}{4}x^2 = 0 \implies \frac{3}{4}x^2 = 27 \implies x^2 = 36 \implies x = 6\text{ cm}.
Stationary points occur where the first derivative of the volume function equals zero.
4
Verify that x=6x = 6 gives a maximum volume and calculate V(6)V(6).
d2Vdx2=−32x  ⟹  d2Vdx2∣x=6=−9<0\frac{d^2V}{dx^2} = -\frac{3}{2}x \implies \left.\frac{d^2V}{dx^2}\right|_{x=6} = -9 < 0 (maximum). Volume V(6)=27(6)−14(63)=162−54=108 cm3V(6) = 27(6) - \frac{1}{4}(6^3) = 162 - 54 = 108\text{ cm}^3.
The negative second derivative confirms a maximum turning point.

Anahtar Kavram

Optimization of physical quantities using the first and second derivative tests.
Tahmini Süre:2m 0s
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Stationary Points, Maxima, and Minima

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