Answer:
Step-by-step explanation:
Find two numbers with a sum of 20 and a difference of 14.
1 answer:
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Answer:
Therefore the dimensions of the square should be 0.1528 inch by 0.1528 inch so, the box has largest volume.
Step-by-step explanation:
Given that,
A box is being created out of a 15 inches by 10 inches sheet of metal.
The length of the one side of the squares which are cut out of the each corners of the metal sheet be x.
The length of the metal box be = (15-2x) inches.
The width of the metal box be =(10-2x) inches
The height of the metal box be =x inches
Then, the volume of the metal box= length×width×height
=(15-2x)(10-2x)x cubic inches
=(150x-50x²+4x³) cubic inches
∴ V= 4x³-50x²+15x
Differentiating with respect to x
V'=12x²-100x+15
Again differentiating with respect to x
V''=24x-100
For maximum or minimum value, V'=0
12x²-100x+15=0
Apply quadratic formula , here a=12, b= -100 and c=15
For x= 8.18, The value of (15-2x) and (10-2x) will negative.
∴x=0.1528 .
Now,
∴At x=0.1528 inch, the volume of the metal box will be maximum.
Therefore the dimensions of the square should be 0.1528 inch by 0.1528 inch so, the box has largest volume.
Look at the graph below carefully
Observe the results of shifting ={2}^{x}f(x)=2x
vertically:
The domain, (−∞,∞) remains unchanged.
When the function is shifted up 3 units to ={2}^{x}+3g(x)=2x +3:
The y-intercept shifts up 3 units to (0,4).
The asymptote shifts up 3 units to y=3y=3.
The range becomes (3,∞).
When the function is shifted down 3 units to ={2}^{x}-3h(x)=2 x −3:
The y-intercept shifts down 3 units to (0,−2).
The asymptote also shifts down 3 units to y=-3y=−3.
The range becomes (−3,∞).
