Find the dimensions of the rectangular box with largest volume if the total surface area is given as 64cm^2
1 answer:
Let x =lenght, y = width, and z =height <span>The volume of the box is equal to V = xyz </span> <span>Subject to the surface area </span> <span>S = 2xy + 2xz + 2yz = 64 </span> <span>= 2(xy + xz + yz) </span> <span>= 2[xy + x(64/xy) + y(64/xy)] </span> <span>S(x,y)= 2(xy + 64/y + 64/x) </span> <span>Then </span> <span>Mx(x, y) = y = 64/x^2 </span> <span>My(x, y) = x = 64/y^2 </span> <span>y^2 = 64/x </span> <span>(64/x^2)^2 = 64 </span> <span>4096/x^4 = 64/x </span> <span>x^3 = 4096/64 </span> <span>x^3 = 64 </span> <span>x = 4 </span> <span>y = 64/x^2 </span> <span>y = 4 </span> <span>z= 64/yx </span> <span>z= 64/16 </span> <span>z = 4 </span> <span>Therefor the dimensions are cube 4.</span>
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Answer:
98°+62°+6x = 180
160+6x = 180
6x = 180 - 160
6x = 20
x= 20/6
x= 3.33
therefor your answer is 3.33
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