{Calculus} Minimize the heat loss

[youngstudent16]

Homework Statement

A box-shaped house has a square floor. Three times as much heat per square foot is lost through the roof as through the walls, but no heat is lost through the floor. If the house encloses 1500 cubic feet, then find the area of the floor that minimizes heat loss.

Homework Equations

First and second derivative test for the absolute extrema

The Attempt at a Solution

I'm very weak with optimization problems, the hardest part is setting up the equation for myself. I figure I want to maximize the size of the floor in order to minimize heat loss. I know my constraint is the volume of the cube. After that I'm kind of stuck.

So Constraint: 1500=lwh
maximize: floor(lw)

Thanks for any help

 

[Chestermiller]

If the dimensions of the house are l, w, and h, what is the area of the roof?
If the dimensions of the house are l, w, and h, what is the area of the walls?

Chet

 

[youngstudent16]

If the dimensions of the house are l, w, and h, what is the area of the roof?
If the dimensions of the house are l, w, and h, what is the area of the walls?

Chet

I think I see what your saying would this help

Say the floor (and roof) is a square of side length x, and the building height is y. Then the area of the roof is x^2, and the total area of the 4 walls, each with area xy, is 4xy

 

[Chestermiller]

I think I see what your saying would this help

Say the floor (and roof) is a square of side length x, and the building height is y. Then the area of the roof is x^2, and the total area of the 4 walls, each with area xy, is 4xy

Good. Now, if the amount of heat lost through the walls is k(4xy), where k is the constant of proportionality between the wall area and the heat loss through the walls, in terms of k, how much heat is lost through the roof? In terms of x, y, and k, what is the total amount of heat loss?

Chet

 

[youngstudent16]

Hmm ok so since you lose 3k through the roof then the total will be
3x^2+4xy
Now I have to figure out how to minimize that, I think I need it as one variable take the derivative and set it equal to 0

 

[Chestermiller]

Hmm ok so since you lose 3k through the roof then the total will be
3x^2+4xy
Now I have to figure out how to minimize that, I think I need it as one variable take the derivative and set it equal to 0

Excellent.

Chet

 

[youngstudent16]

Ok x^2y=1500
I solved for x and then substitued that into the equation above

took the derivative set it equal to 0 to solve for y as 15

plugged y into x^2y=1500 to get x = 10

Would then the area of the floor that minimizes the heat loss be 3(10)^2

or 300

 

[SteamKing]

Hmm ok so since you lose 3k through the roof then the total will be
3x^2+4xy
Now I have to figure out how to minimize that, I think I need it as one variable take the derivative and set it equal to 0

You're forgetting about expressing the volume of the house in terms of x and y.

Adding this constraint can help simplify the equation for the heat loss thru the walls and roof.

(Instead of jotting down expressions, try to write complete equations for the heat loss.)

 

[Chestermiller]

Ok x^2y=1500
I solved for x and then substitued that into the equation above

took the derivative set it equal to 0 to solve for y as 15

plugged y into x^2y=1500 to get x = 10

Would then the area of the floor that minimizes the heat loss be 3(10)^2

or 300

Good work, but no, it would be 100. The area is 10 x 10.

Chet

 

[youngstudent16]

Good work, but no, it would be 100. The area is 10 x 10.

Chet

oops yes thank you for the help