Let h = height of the box,
x = side length of the base.
Volume of the box is

.
So

Surface area of a box is S = 2(Width • Length + Length • Height + Height • Width).
So surface area of the box is


The surface are is supposed to be the minimum. So we'll need to find the first derivative of the surface area function and set it to zero.

![4x = \frac{460}{ x^{2} } \\ 4x^{3} = 460 \\ x^{3} = 115 \\ x = \sqrt[3]{115} = 4.86](https://tex.z-dn.net/?f=%204x%20%3D%20%5Cfrac%7B460%7D%7B%20x%5E%7B2%7D%20%7D%20%20%5C%5C%20%204x%5E%7B3%7D%20%3D%20460%20%20%5C%5C%20x%5E%7B3%7D%20%3D%20115%20%20%5C%5C%20x%20%3D%20%20%5Csqrt%5B3%5D%7B115%7D%20%3D%204.86%20)
Then

So the box is 4.86 in. wide and 4.87 in. high.
<span>-4x + y = -25 y = 4x - 25
-6x - 6y = 0 </span>
-6x - 6(4x - 25) = 0
-6x - 24x + 150 = 0
-30x + 150 = 0
-30x = -150
x = 5
y = 4(5) - 25
y = 20 - 25
y = -5
(5, -5)
Let x be the number
5x = x + 5
subtract x from both sides
5x - x = 5
4x = 5
Divide both sides by 4
4x/4 = 5/4
x = 5/4
=======
Check:
5 * (5/4) = (5/4) + 5
25/4 = 1.25 + 5
6.25 = 6.25
Hope it helps . Note that I am a sixth grade student and I don’t know anything . I got help from somebody
Answer:1.2 m/s
Step-by-step explanation:
Total distance covered is 500+500+200 m so the average velocity is given by
V = distance/time
V=1200/(17*60)
V=1.176 m/sec