What is the Smallest Surface Area of a Crate Delivered by Canada Post?

In summary, the conversation discusses the maximum size restrictions for parcels that Canada Post will deliver and the smallest surface area of a crate that meets these restrictions. The equation for girth of a rectangular prism is given and two methods for finding the smallest surface area are suggested: setting up and solving equations using the given constraints, or using the Lagrange Multiplier method.
  • #1
S.R
81
0

Homework Statement


Canada Post will deliver parcels only if they are less than a certain maximum size: the combined length and girth cannot exceed 297 cm (Girth is the total distance around the cross-section of the parcel). Canada Post delivers a crate with the smallest SA to your house. What is the SA of the crate in square meters?

Homework Equations


Girth of a rectangular prism=2(w+h) -> Web-search
I'm still unclear on what a girth is, however. Maybe it is 2(l+w), the perimeter of the base.

The Attempt at a Solution


I set-up the equations:

l+2(w+h)=297

SA=2(wh+lw+lh)

I'm unsure how to proceed. Any help is appreciated. Thanks!

S.R
 
Last edited:
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  • #2
S.R said:

Homework Statement


Canada Post will deliver parcels only if they are less than a certain maximum size: the combined length and girth cannot exceed 297 cm (Girth is the total distance around the cross-section of the parcel). Canada Post delivers a crate with the smallest SA to your house. What is the SA of the crate in square meters?
The smallest? You say that Canada Post will not deliver packages above a certain size but they can be as small as you please. Are you asking for the smallest surface are of a package that meets the maximum sum of length and girth?

Homework Equations


Girth of a rectangular prism=2(w+h) -> Web-search
I'm still unclear on what a girth is, however. Maybe it is 2(l+w), the perimeter of the base.
Yes. assuming you are taking h as the longest side, "girth" is 2(l+ w) so the requirement is that [itex]h+ 2(l+w)\le 297[/itex].

The Attempt at a Solution


I set-up the equations:

l+2(w+h)=297
Hey, you switched h and l on me!

SA=2(wh+lw+lh)

I'm unsure how to proceed. Any help is appreciated. Thanks!

S.R
Proceed in either of two ways:
1) use l+ 2*(w+ h)= 297 (or h+ 2(w+ l)= 297) to eliminate one of the three variables leaving only two. Set the partial derivatives with respect to the two variables equal to 0 and solve the two equations.

2) Use the "Lagrange Multiplier" method. Form the gradient of the "object function", 2(wh+ lw+ lh), the gradient of the constraint, l+ 2(w+h), and set one equal to a constant ([itex]\lambda[/itex] times the other.
 
  • #3
HallsofIvy said:
Proceed in either of two ways:
1) use l+2*(w+ h)= 297 (or h+ 2(w+ l)= 297) to eliminate one of the three variables leaving only two. Set the partial derivatives with respect to the two variables equal to 0 and solve the two equations.

I'm not sure how to eliminate one of the three variables?
 
Last edited:

Related to What is the Smallest Surface Area of a Crate Delivered by Canada Post?

1. What are the basic properties of quadratic equations?

The basic properties of quadratic equations include having a degree of 2, having a parabolic shape, and having two solutions (roots) when graphed on a coordinate plane.

2. How do you determine the number of solutions for a quadratic equation?

The number of solutions for a quadratic equation can be determined by the discriminant, which is the expression under the square root in the quadratic formula. If the discriminant is positive, there are two real solutions. If the discriminant is zero, there is one real solution. If the discriminant is negative, there are no real solutions.

3. What is the difference between the vertex form and standard form of a quadratic equation?

The vertex form of a quadratic equation is written as y = a(x-h)^2 + k, where (h,k) represents the coordinates of the vertex. The standard form is written as y = ax^2 + bx + c, where a, b, and c are constants. The vertex form is useful for finding the vertex and making transformations, while the standard form is useful for graphing and solving for x-intercepts.

4. How do you solve a quadratic equation by factoring?

To solve a quadratic equation by factoring, you must first rearrange the equation so that it is equal to zero. Then, factor the quadratic expression into two binomials. Finally, set each binomial equal to zero and solve for the variable. The solutions will be the x-intercepts of the quadratic curve.

5. Can a quadratic equation have imaginary solutions?

Yes, a quadratic equation can have imaginary solutions. This occurs when the discriminant is negative, meaning that the solutions are complex numbers. Complex solutions are often represented as a+bi, where a and b are real numbers and i represents the imaginary unit. In this case, the quadratic equation will not have any x-intercepts when graphed on a coordinate plane.

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