I believe <span>(4, –2) is correct :)</span>
Step-by-step explanation:
what's the question about
Answer:
(a)Length =2 feet
(b)Width =2 feet
(c)Height=3 feet
Step-by-step explanation:
Let the dimensions of the box be x, y and z
The rectangular box has a square base.
Therefore, Volume of the box
Volume of the box
The material for the base costs 
, the material for the sides costs 
, and the material for the top costs 
.
Area of the base 
Cost of the Base 
Area of the sides 
Cost of the sides=
Area of the Top
Cost of the Base 
Total Cost, 
Substituting 
To minimize C(x), we solve for the derivative and obtain its critical point
![C'(x)=\dfrac{0.6x^3-4.8}{x^2}\\Setting \:C'(x)=0\\0.6x^3-4.8=0\\0.6x^3=4.8\\x^3=4.8\div 0.6\\x^3=8\\x=\sqrt[3]{8}=2](https://tex.z-dn.net/?f=C%27%28x%29%3D%5Cdfrac%7B0.6x%5E3-4.8%7D%7Bx%5E2%7D%5C%5CSetting%20%5C%3AC%27%28x%29%3D0%5C%5C0.6x%5E3-4.8%3D0%5C%5C0.6x%5E3%3D4.8%5C%5Cx%5E3%3D4.8%5Cdiv%200.6%5C%5Cx%5E3%3D8%5C%5Cx%3D%5Csqrt%5B3%5D%7B8%7D%3D2)
Recall: 
Therefore, the dimensions that minimizes the cost of the box are:
(a)Length =2 feet
(b)Width =2 feet
(c)Height=3 feet
So, pretend this is your x-axis and y-axis:
I
I
(-2,7) • I
I
I • (2, 5)
I
I
I
I
_________________I____________________
I
I
I
TO GET FROM POINT (-2, 7) TO POINT (2, 5), WE MOVE DOWN 2 AND OVER 4, SO THE SLOPE IS -1/2. IF WE FOLLOW THAT SLOPE AND MOVE DOWN 1 AND OVER 2 FROM THE FIRST POINT OF (-2, 7), WE WILL LAND ON A POINT LOCATED AT (0, 6), WHICH WOULD BE THE "Y-INTERCEPT". WE WERE JUST ABLE TO CALCULATE THE SLOPE OF THE LINE AND THEN USE THE SLOPE TO FIND THE INTERCEPT. SO, THE "SLOPE-INTERCEPT" FORM OF THE EQUATION FOR THIS LINE IS:
y = -1/2x + 6
TO RE-WRITE THIS IN STANDARD FORM, WE JUST WANT TO MOVE THE X VARIABLE OVER TO THE LEFT WITH THE Y VARIABLE, SO:
y = -1/2x + 6
+1/2x + 1/2x
1/2x + y = 6 .... and that is your answer!
