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Exercise 7.8 · Q10

Q.A manufacturer wants to design an open box having a square base and a surface area of 108,textsq.cm108\\,\\text{sq.cm}. Determine the dimensions of the box for the maximum volume.

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Use the fixed open-top surface area to express height in terms of the base side, reduce volume to one variable, then maximize.

Step 1. Set up. Let the square base have side xx and the box have height hh. Open-top surface area: S=x2+4xh=108⇒h=108−x24xS=x^2+4xh=108\Rightarrow h=\dfrac{108-x^2}{4x}.

V(x)=x2h=x2⋅108−x24x=108x−x34.V(x)=x^2h=x^2\cdot\frac{108-x^2}{4x}=\frac{108x-x^3}{4}.

Step 2. Differentiate and solve V′(x)=0V'(x)=0.

V′(x)=108−3x24=0 ⇒ x2=36 ⇒ x=6 (x>0).V'(x)=\frac{108-3x^2}{4}=0\ \Rightarrow\ x^2=36\ \Rightarrow\ x=6\ (x>0).

Step 3. Confirm maximum. …

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