Final answer:
To find the dimensions of the bag with the largest volume, we can use optimization. By setting up equations and constraints, we can find the values of the dimensions that maximize the volume. The bag with the largest volume that Fly-by-night airlines will accept measures approximately 57 inches in length, 57 inches in width, and 57/2 inches in height.
Step-by-step explanation:
To find the dimensions of the bag with the largest volume that Fly-by-night airlines will accept, we can use the concept of optimization. Let's assume the length of the bag is l, the width is w, and the height is h. According to the rules, l + w ≤ 57 and w + h ≤ 57.
The volume of the bag is V = lwh. We want to maximize V subject to the constraints l + w ≤ 57 and w + h ≤ 57.
One way to solve this problem is by using calculus. We can set up an equation to express V in terms of a single variable. From the constraint l + w ≤ 57, we can rewrite it as l = 57 - w. Substituting this into the volume equation, we have V = (57 - w)wh = 57w - w^2h.
To find the maximum volume, we can take the derivative of V with respect to w and set it equal to 0. Differentiating V with respect to w, we get dV/dw = 57 - 2wh. Setting this equal to 0, we have 57 - 2wh = 0. Solving for w, we get w = 57/(2h).
Substituting this value of w back into the volume equation, we have V = 57w - w^2h = 57(57/(2h)) - (57/(2h))^2h = (57^2)/(2h) - (57^2)/(4h) = (57^2)/(4h).
Now, we can maximize V by finding the value of h that minimizes the denominator (4h). Since h is constrained by w + h ≤ 57, we know that h is at most 57 - w. To maximize V, we want to minimize the denominator, so we choose the largest value of h that satisfies this constraint, which is h = 57 - w.
Substituting this value of h back into the volume equation, we have V = (57^2)/(4h) = (57^2)/(4(57 - w)) = (57^2)/(228 - 4w).
To find the maximum volume, we can take the derivative of V with respect to w and set it equal to 0. Differentiating V with respect to w, we get dV/dw = - (57^2)/(228 - 4w)^2(-4) = (57^2)(4)/(228 - 4w)^2. Setting this equal to 0, we have (57^2)(4)/(228 - 4w)^2 = 0. Solving for w, we get w = 57/2.
Substituting this value of w back into the volume equation, we have V = (57^2)/(4(57 - w)) = (57^2)/(4(57 - 57/2)) = 29,796.75 inches cubed.