Functions of Several Variables — Question 10

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Question 10

An open-top box has length xx, width yy, and height hh. A design constraint fixes the sum of these dimensions: x+y+h=12,x,y,h>0.x+y+h=12,\qquad x,y,h>0. Its volume is V(x,y)=xy(12−x−y)V(x,y)=xy(12-x-y).

Tasks

  1. State the natural domain.

  2. Find the greatest possible volume without calculus.

  3. Prove the maximizing dimensions are unique.

Original worksheet page 1: question and worked solution for 1-5-010
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Question 10 – Solution

Strategy Interpret the three positive factors xx, yy, and h=12−x−yh=12-x-y, then apply AM–GM.

See the diagram in the original worksheet below.

Step 1: Domain Positivity requires x>0,y>0,x+y<12.\boxed{x>0,\qquad y>0,\qquad x+y<12}. This is the interior of a triangular region.

Step 2: Bound The three positive numbers x,y,hx,y,h have fixed sum 1212. AM–GM gives x+y+h3≥xyh3⇒4≥V3.\frac{x+y+h}{3}\ge\sqrt[3]{xyh}\Longrightarrow 4\ge\sqrt[3]{V}. Cubing yields V≤64V\le 64.

Step 3: Equality Equality in AM–GM occurs exactly when x=y=h=4.x=y=h=4. Thus the unique maximizing dimensions are x=4,y=4,h=4\boxed{x=4,y=4,h=4} and Vmax=64.\boxed{V_{\max}=64}. Direct substitution satisfies the material constraint and confirms the volume.

Original worksheet page 2: question and worked solution for 1-5-010

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