. A rectangle has an area of 16 square feet. Its length and width are whole numbers. What is its maximum perimeter?

(L) x (W) = 16 2L + 2W = P Solve for one of the variables, in this case lets solve for (w) LW = 16 W = 16/L plug that in to the Perimeter Eqauation: 2(L) + 2( 16/L ) = P Now take the derivative of P P = 2L + 32/L P'= 2 + 32( -1/L^2 ) Now set P' to 0, because the derivative is 0 at its Maximums and Minimums 0 = 2 - 32/(L^2) -2 = -32/(L^2) -2(L^2) = -32 L^2 = 16 L = 4 Plug this back in to solve for W W = 16/L W = 16/4 W = 4

P = 2x + 2y, A = xy =>P = 2x + 2A/x dP/dx = 0 for max. find value of x where that occurs.

so the maximum perimeter must be 16, and the dimensions are 4 by 4

@heavenfun13 you should read the code of conduct and help students do the work themselves, not do it for them

i did

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