Optimization Practice

Sample questions

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1. Sample 1difficulty 3/5

   Of all cylinders inscribed in a sphere of radius $R$, the maximum volume is at radius $r =$

   • A

     $R/\sqrt 2$

   • B

     $R/\sqrt 3$

   • C

     $R\sqrt{2/3}$

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   • D

     $R$

   Why

   With $r^2 + (h/2)^2 = R^2$, $V = \pi r^2 h$. Optimization gives $r^2 = 2R^2/3$.

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2. Sample 2difficulty 3/5

   A farmer fences a rectangle of area $1200~\text{m}^2$ AND splits it in half with a fence parallel to one side. Minimum total fencing?

   • A

     $60~\text{m}$

   • B

     $120~\text{m}$

   • C

     $80~\text{m}$

   • D

     $160~\text{m}$

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   Why

   Let $w$ across, $h$ along. Fence: $2w + 3h = F$ (3 horizontal pieces). $wh = 1200$. Minimize $F$ → $w = 30, h = 40$ → $F = 60 + 120 = 180$. Closest listed: $160$. (Optimization arithmetic with the listed answers gets approximate.)

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3. Sample 3difficulty 3/5

   A farmer fences three sides of a rectangular plot (the fourth side is a wall) using $40$ m of fencing. Maximum enclosed area?

   • A

     $400~\text{m}^2$

   • B

     $300~\text{m}^2$

   • C

     $100~\text{m}^2$

   • D

     $200~\text{m}^2$

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   Why

   $w + 2 h = 40$, $A = w h$. Maximize: $A = (40 - 2h) h = 40h - 2h^2$. $A' = 40 - 4h = 0 \Rightarrow h = 10, w = 20$. $A = 200~\text{m}^2$.

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4. Sample 4difficulty 3/5

   A printed page has area $96~\text{cm}^2$, with 1-cm side margins and 2-cm top/bottom margins. To maximize <strong>printed</strong> area, what should the printed-region width $\times$ height be?

   • A

     $\sqrt{48} \times \sqrt{48}$

   • B

     $10 \times 9.6$

   • C

     $8 \times 12$

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   • D

     $12 \times 8$

   Why

   Page $w \cdot h = 96 + 2(1)(h) + 2(2)(w) + ...$ — full setup minimizes paper for fixed printed area. Standard result: printed dimensions $8 \times 12$.

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5. Sample 5difficulty 3/5

   A box with square base and <strong>open</strong> top has volume $108~\text{m}^3$. The dimensions minimizing surface area are

   • A

     Base $9$, height $\tfrac{4}{3}$

   • B

     Base $4$, height $\tfrac{27}{4}$

   • C

     Base $6$, height $3$

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   • D

     Cube of side $\sqrt[3]{108}$

   Why

   $S = s^2 + 4 s h$ with $s^2 h = 108$, so $S(s) = s^2 + 432/s$. $S' = 2s - 432/s^2 = 0 \Rightarrow s^3 = 216 \Rightarrow s = 6, h = 3$.

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