minimum area problem?

From the first equation 27 = L*W you get W = 27/L to substitute in the other.

A = (L + 3)(27/L + 4)

Multiply out, find dA/dL and equate to zero.

Can you take it from there?

P.S. You appear to be using L for the width of the page. I would put

A = (L + 4)(27/L + 3)

27 = L*W

so L = 27/W

and also, A = (L+4)*(W+3)

sub the first eqn:

A = (27/W + 4)*(W+3)

A = 27 + 4W + 81/W + 12

A = 4w + 81W^(-1) + 39

to find min size, make A'=0

A' = 4 -81W^(-2)

0 = 4 - 81/W^2

81/W^2 = 4

81 = 4W^2

20.25 = W^2

9/2 = W

so where the width is 9/2 and the length is:

L = 27/(9/2)

L = 6

so W is 4.5 and L is 6 would be the least paper.

Ha but you know 27 = L * W, so L = 27 / W.

A = (W + 3) * (L + 4) (note my correction to yours)

so A = (W + 3) * (27/W + 4)

A = 27 + 12 + 81 W^(-1) + 4W

dA / dW = -81W^(-2) + 4 = 0

so 81 / W^(2) = 4

so W^2 = 81/4

so W = 9/2

quantity of an rectangular prism: V=lwh; V=22 opportunities(lengths in random order): 22x1x1 2x11x1 (those are the only opportunities i could desire to make your concepts up, there is probably greater, yet could likely incorporate extremely some decimals.) floor component of an rectangular prism: A=2lw+2wh+2lh First risk:2x11x1 A=2(2)(11)+2(11)(a million)+2(2)(a million) A=40 4+22+4 A=70 cm² 2d risk: 22x1x1 A=2(22)(a million)+2(a million)(a million)+2(22)(a million) A=40 4+2+40 4 A=ninety cm² answer: 2x11x1