Answer:
Explanation:
I think your question is missed of key information, allow me to add in and hope it will fit the original one. Please have a look at the attached photo.
Given:
Let X is the the length
Let Y is the width
From the photo, we can see that
(4X + 6Y)*71 = 34080
<=> (4X + 6Y) = 480
<=> Y = 80 - [tex]\frac{2}{3}[/tex]X
The are of the rectangular industrial warehouse:
A(X) = 3Y*X
<=> A(X) = 3(80 - [tex]\frac{2}{3}[/tex]X )X
<=>A(X) = (240-2X)X = 240X - [tex]2X^{2}[/tex]
So A'(X) = 240 - 4X
Let A'(X) = 0, we have:
240 - 4X = 0
<=> X = 60
=> Y =(80 - [tex]\frac{2}{3}[/tex]X ) = 80 - [tex]\frac{2}{3}[/tex]*60 = 40
So the dimension to maximize total area is: 60 in length and 40 in width