CALCULUS - Problem 01 Find the shape of the rectangle of maximum perimeter inscribed in a circle Solution 01 Perimeter of rectangle P=2x 2y Where x=Dcos

# CALCULUS - Problem 01 Find the shape of the rectangle of...

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Problem 01 Find the shape of the rectangle of maximum perimeter inscribed in a circle. Solution 01 Perimeter of rectangle, P=2x+2y Where: x=Dcosθ y=Dsinθ P=2Dcosθ+2Dsinθ dPdθ=−2Dsinθ+2Dcosθ=0 −sinθ+cosθ=0 sinθ=cosθ sinθcosθ=1 tanθ=1 θ=45 x=Dcos45 =0.707D y=Dsin45 =0.707D x=y (square) answer Solution 02 Convex surface area of cylinder, A=πdh Where: d=Dcosθ h=Dsinθ A=π(Dcosθ)(Dsinθ) A=D2πcosθsinθ dAdθ=D2π(cos2θ−sin2θ)=0 sin2θ=cos2θ tan2θ=1 θ=45 d=Dcos45 =0.707D h=Dsin45 =0.707D diameter=height answer Problem 3 Find the weight of the heaviest circular cylinder can be cut from a 16-lb shot. Solution 03 A shot is in the form of a sphere and the cylinder is the cylinder of maximum. From the figure )This is also the figure used in Solution 02 : Vc=14πd2h Where: d=Dcosθ h=Dsinθ Thus, Vc=14π(Dcosθ)2(sinθ) Vc=14D3πcos2θsinθ dVcdθ=14D3π[cos2θ(cosθ)+sinθ(−2cosθsinθ)] dVcdθ=14D3π(cos3θ−2sin2θcosθ)=0 2sin2θcosθ=cos3θ 2sin2θ=cos2θ sin2θcos2θ=12 tan2θ=12 tanθ=12√ cosθ=2√3√ sinθ=13√ Vc=14D3π(2√3√)2(13√)
Vc=163√πD3 → Maximum volume of cylinder Volume of shot (sphere): Vs=43πR3=43π(D/2)3 Vs=16πD3 Weight is proportional to the volume, so WcVc=WsVs Wc=WsVs×Vc Wc=1616πD3×163√πD3 Wc=9.24 lb answe r Problem 4 The stiffness of a rectangular beam is proportional to the breadth and the cube of the depth. Find the shape of the stiffest beam that can be cut from a log of a given size. Solution 4 Stiffness, k=bd3 Where: b=Dcosθ d=Dsinθ k=D4cosθsin3θ dkdθ=D4(3cos2θsin2θ−sin4θ)=0 3cos2θ−sin2θ=0 sin2θ=3cos2θ tan2θ=3 tanθ=3√ θ=60 b=Dcos60 =12D d=Dsin60 =123√D depth=3√×breadth answer Problem 5 The strength of rectangular beam is proportional to the breadth and the square of the depth. Find the shape of the strongest beam that can be cut from a log of given size.