Answer:
Step-by-step explanation: this is easy dubs
Answer:
hypotenuse²=base²+height²
![\sqrt{85 } {}^{2} = 6 {}^{2} + x {}^{2} \\](https://tex.z-dn.net/?f=%20%5Csqrt%7B85%20%7D%20%7B%7D%5E%7B2%7D%20%20%20%3D%206%20%7B%7D%5E%7B2%7D%20%20%2B%20x%20%7B%7D%5E%7B2%7D%20%20%5C%5C%20)
<h2>85=36+x²</h2><h2>85-36=x²</h2><h2>49=x²</h2><h2>49½=x</h2><h2>x=7 </h2>
Answer:
137473735858884506343
Step-by-step explanation:
Answer:
x = 2.354 in
Maximum volume = 228.162 in³
Step-by-step explanation:
Size of the metal sheet = (18 in × 12 in)
Let the height of the box = x in
Then length of the box = (18 - 2x) in
Width of the box = (12 - 2x) in
Now volume of the rectangular box (V) = Length × Width × height
= (18 - 2x) × (12 - 2x) × x
= x[18(12 - 2x) - 2x(12 - 2x)]
= x[216 - 36x - 24x + 4x²]
= x[216 - 60x + 4x²]
= (4x³ - 60x² + 216x) in³
For maximum volume, we will find the derivative of the volume and equate it to zero.
![\frac{dV}{dx}=\frac{d}{dx}(4x^3-60x^2+216x)](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdx%7D%3D%5Cfrac%7Bd%7D%7Bdx%7D%284x%5E3-60x%5E2%2B216x%29)
V' = 12x² - 120x + 216
V' = 0
12x² - 120x + 216 = 0
x² - 10x + 18 = 0
By quadratic formula,
x = ![\frac{10\pm\sqrt{(-10)^2-4(1)(18)}}{2(1)}](https://tex.z-dn.net/?f=%5Cfrac%7B10%5Cpm%5Csqrt%7B%28-10%29%5E2-4%281%29%2818%29%7D%7D%7B2%281%29%7D)
x = 5 ± √7
x = 7.646, 2.354
But for 7.646, volume of the box will be negative.
Therefore, (x = 2.354 in) will be value of x for maximum volume of the box.
Maximum volume of the box = 4(2.354)³ - 60(2.354)² + 216(2.354)
= 228.162 in³