Find the dimensions of the rectangle of maximum area whose perimeter is 16 cm ?
The rectangle of maximum area is a square of side length
Then the perimeter is:
so that:
The area is then:
Find the critical points of the function:
and as:
the critical points is a maximum.
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Let the length of the rectangle be ( l ) cm and the width be ( w ) cm. Given that the perimeter is 16 cm, we have the equation:
[ 2l + 2w = 16 ]
Solving this equation for ( l ), we get:
[ l = 8 - w ]
The area of the rectangle, ( A ), is given by:
[ A = lw ]
Substituting ( l = 8 - w ) into the area formula, we get:
[ A = (8 - w)w ]
To find the maximum area, we can take the derivative of ( A ) with respect to ( w ), set it equal to zero, and solve for ( w ):
[ \frac{dA}{dw} = 8 - 2w ] [ 8 - 2w = 0 ] [ w = 4 ]
Substituting ( w = 4 ) back into the equation for ( l ), we find:
[ l = 8 - 4 = 4 ]
Therefore, the dimensions of the rectangle of maximum area with a perimeter of 16 cm are ( 4 ) cm by ( 4 ) cm.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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