Aposter of area 19,440 cm2 has blank margins of width 10 cm on the top and bottom and 6 cm on the sides. find the dimensions that maximize the printed area. (let w be the width of the poster, and let h be the height.)

h = 180 cm w = 108 cm First, let's express width (W) as a function of height (H). 19440 = WH 19440 / H = W Now let's express printed area using W and H A = (W - 2*6)(H - 2*10) A = (W - 12)(H - 20) Now substitute the equation for W into the formula A = (W - 12)(H - 20) A = (19440/H - 12)(H - 20) And expand and simplify A = (19440/H - 12)(H - 20) A = H*19440/H - 12H - 20*19440/H + 240 A = 19440 - 12H - 388800/H + 240 A = 19680 - 12H - 388800/H Since we're looking for a maximum, that will happen when the slope of the function is 0. And the first derivative of the function will calculate the slope of each point of the function. So let's calculate the first derivative. A = 19680 - 12H - 388800/H A = 19680*H^0 - 12H - 388800*H^(-1) A' = 0*19680*H^(0-1) - 1*12H^(1-1) - (-1)*388800*H^(-1-1) A' = 0*19680*H^(-1) - 1*12H^(0) - (-1)*388800*H^(-2) A' = 0*19680/H - 1*12*1 - (-1)*388800/H^2 A' = 0 - 12 + 388800/H^2 A' = 388800/H^2 - 12 Now set to 0 and solve for H A' = 388800/H^2 - 12 0 = 388800/H^2 - 12 12 = 388800/H^2 12H^2 = 388800 H^2 = 388800/12 = 32400 H = 180 So the optimal height is 180 cm. The width will be 19440 / 180 = 108 cm

Sis less than or equal to 75 (bottom left)