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Expert: Scott A Wilson - 11/7/2009
Question
1. Find the two real numbers whose sum is 18 and whose product is a minimum.
2. Find the two real numbers whose sum is 24 and whose product is a maximum.
3. The perimeter of a rectangle is 38m. Find the dimensions of the rectangle that will contain the greatest area.
4. The longer leg of a right triangle is 2cm more than twice the shorter leg. Find the measure of each leg of the triangle with the smallest possible area.
5. Alex has 20m of barb wire to fence a rectangular lot. What are the dimensions of the lot that will contain the maximum area?
6. A cellular phone store can sell 1000 phone kits per month for P5000 each. Research indicates that the store can sell 50 more kits per month for each P100 decrease in price. What price per kit will give the greatest monthly sales? What is the sales amount?
I hope you can answer this one. Thank you.
I'm from the PH btw. =)
Answer
1. x+y=18, so y=18-x; maximize f(x) = x(19-x) = 19x - x² by solving f'(x) = 0.
2. x+y=24, so y=24-x; maximize f(x) = x(24-x) = 24x - x² by setting f'(x) = 0.
3. Perimeter = 2W + 2L = 38cm so W+L=19, W=19-L;
put that in area A = WL and get A=(19-L)L, so A=19L-L².
4. Two sides a right triangle are L and S, and L = 2 + 2S = 2(1+S), all in centimeters.
Area A = LS/S, substitute L = 2(1+S) in and get A = (1+S)S, so A(S) = S + S².
Find A'(S), set to 0, solve for S, add 1, multiply by 2, and you have L too.
5. For Alex, let the lot be the sides be L and W.
The total fences is 2L + 2W = 20, all in meters.
This means that L + W = 10, so L = 10 - W.
Since the area A = LW, put in L and get A(W) = (10-W)W = 10W - W².
Set A'(W) = 0, solve for W, rememer L = 10 - W, and there you have L and W.
6. N = 1000 phone kits, P = 5000 each, cash C = NP.
Research indicates that N has 50 more for each 100 decrease in P.
That is an increas in N of 1 for each decrease in P of 2.
This says we should take C = (N+x)(P-2x) = NP + xP - 2Nx - 2x².
Thus, if C(x) = NP + xP - 2Nx - 2x², C'(x) = P - 2N - 4x. This gives x = (2N-P)/4.
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