William W. answered • 07/22/20
Experienced Tutor and Retired Engineer
Since we are being asked to find the smallest area, we know that we are trying to find a minimum of the area function.
Let's let "W" be the overall width of poster and "H" be the overall height of the poster.
A = H•W
But we also have information that will help us get the number of variables down to one. We know that the area of the printed material is 384. The width of the printed material is W - 4(2) or W - 8. The height of the printed material is H - 6(2) = H - 12. So the area of the printed material is (W - 8)(H - 12) and that equals 384
(W - 8)(H - 12) = 384
WH - 12W - 8H + 96 = 384
W(H - 12) - 8H = 288
W = (288 + 8H)/(H - 12)
Plugging that into the area equation (A = H•W), we get:
A(H) = H(288 + 8H)/(H - 12)
A(H) = (288H + 8H2)/(H - 12)
To find the minimum, take the derivative and set it equal to zero.
To take the derivative, we must use the quotient rule
If f(x) = u/v then f '(x) = (u'v - uv')/v2
In this case u = 288H + 8H2 so u' = 288 + 16H
and v = H - 12 so v' = 1 and v2 = (H - 12)2 so
A'(H) = [(288 + 16H)(H - 12) - (288H + 8H2)(1)]/(H - 12)2
A'(H) = (288H - 3456 + 16H2 - 192H - 288H - 8H2)/(H - 12)2
A'(H) = (8H2 - 192H - 3456)/(H - 12)2
A'(H) = 8(H2 - 24H - 432)/(H - 12)2
Setting it equal to zero:
8(H2 - 24H - 432)/(H - 12)2 = 0
(H2 - 24H - 432)/(H - 12)2 = 0 This can only be zero if the numerator is zero so:
H2 - 24H - 432 = 0
(H - 36)(H + 12) = 0
H = 36 or H = -12
We can throw out the negative answer since the poster can't be a negative height so H = 36. Since H = 36, and we know that W = (288 + 8H)/(H - 12), we can solve for W:
W = (288 + 8(36))/(36 - 12)
W = 24
So the poster has
W = 24 cm
H = 36 cm
Jeff K.
Sorry, the system removed all the line breaks. -:-( Hope you can still read it.07/22/20
Jeff K.
Here's how to do it without solving any quadratics! Using the same notation: (W-8)(H-12) = 384 Which means: W = 384/(H-12) + 8 . . . . . . . eqn (1) So, total area = A = WH = H [384/((H-12) + 8] By the chain rule, dA/dH = 1.[(384/(H-12) +8] + H[-384/(H-12)^2] For min/max area, dA/dH = 0 => 384/(H-12) + 8 - 384H/(H-12)^2 = 0 Common denominator is (H-12)^2: 384(H-12) + 8(H-12)^2 - 384H = 0 We can multiply both sides by the denominator, (H-12)^2, since it can't be zero. Can you see why from the original diagram? Therefore: 384H - 384 x 12 + 8(H - 12)^2 - 384H = 0 The first and last terms on the left cancel to zero, so we are left with: 8(H - 12)^2 = 384 x 12 (H - 12)^2 = 384 x 12 / 8 (H - 12)^2 = 48 x 12 (H - 12) = sqrt(48 x 12) = 24 . . . . . [taking sq roots on both sides Therefore, H = 12 + 24 = 36 Substitute into eqn (1) above to get W = 16 + 8 = 24 Solution set: (W, H) = (24, 36)07/22/20