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: m = –3±√ 5
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−b±√b2−4ac
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m=
−(6)±√(6)2−4(1)(4)
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m=
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m=−3+√5 or m=−3−√5
